Differentiation: past paper questions & worked solutions

$y=2x^{5/2}-4x+3$, so $\dfrac{dy}{dx}=2\cdot\dfrac52x^{3/2}-4=5x^{3/2}-4$.

At $P$, the gradient is $16$, so $5x^{3/2}-4=16$.

$5x^{3/2}=20$, hence $x^{3/2}=4$.

Since the x-coordinate of $P$ is $2^k$, we have $(2^k)^{3/2}=4=2^2$.

$2^{3k/2}=2^2$, so $\dfrac{3k}{2}=2$ and $k=\dfrac43$.

$y=5x^3+3x^{-2}-7x$.

$\dfrac{dy}{dx}=15x^2-6x^{-3}-7=15x^2-\dfrac{6}{x^3}-7$.

$\dfrac{d^2y}{dx^2}=30x+18x^{-4}=30x+\dfrac{18}{x^4}$.

$y=2+10x^{1/2}-2x^{3/2}$, so $\dfrac{dy}{dx}=5x^{-1/2}-3x^{1/2}$.

$\dfrac{dy}{dx}=\dfrac{5}{\sqrt{x}}-3\sqrt{x}$.

At $x=2$, the gradient is $\dfrac5{\sqrt2}-3\sqrt2=\dfrac5{\sqrt2}-\dfrac6{\sqrt2}=-\dfrac1{\sqrt2}$.

$-\dfrac1{\sqrt2}=-\dfrac{\sqrt2}{2}$.

For $y=3x^5+4x^3-x+5$, $\dfrac{dy}{dx}=15x^4+12x^2-1$.

$15x^4+12x^2-1=2$, so $15x^4+12x^2-3=0$.

Dividing by $3$ gives $5x^4+4x^2-1=0$. Let $u=x^2$, giving $5u^2+4u-1=0$.

$5u^2+4u-1=(5u-1)(u+1)=0$. Since $u=x^2\ge0$, $u=\dfrac15$.

$x^2=\dfrac15$, so $x=\pm\dfrac1{\sqrt5}=\pm\dfrac{\sqrt5}{5}$.

$y=4x^3+\dfrac{3}{2}x^{-1/2}-8$.

$\dfrac{dy}{dx}=12x^2+\dfrac32\left(-\dfrac12\right)x^{-3/2}$.

$\dfrac{dy}{dx}=12x^2-\dfrac34x^{-3/2}=12x^2-\dfrac{3}{4x^{3/2}}$.

$\int y\,dx=\int\left(4x^3+\dfrac32x^{-1/2}-8\right)dx$.

$\int y\,dx=x^4+3x^{1/2}-8x+c=x^4+3\sqrt{x}-8x+c$.

$y=6x^2+3x^{1/2}+\dfrac58$.

$\dfrac{dy}{dx}=12x+\dfrac32x^{-1/2}=12x+\dfrac{3}{2\sqrt{x}}$.

At $x=\dfrac14$, $\dfrac{dy}{dx}=12\left(\dfrac14\right)+\dfrac32\left(\dfrac14\right)^{-1/2}=3+3=6$.

The tangent passes through $P\left(\dfrac14,\dfrac52\right)$, so $y-\dfrac52=6\left(x-\dfrac14\right)$.

$y-\dfrac52=6x-\dfrac32$, hence $y=6x+1$.

$y=\dfrac{x^3}{4}-x^2+\dfrac{17}{x}=\dfrac14x^3-x^2+17x^{-1}$.

$\dfrac{dy}{dx}=\dfrac34x^2-2x-17x^{-2}=\dfrac34x^2-2x-\dfrac{17}{x^2}$.

At $x=2$, the tangent gradient is $\dfrac34(4)-2(2)-\dfrac{17}{4}=3-4-\dfrac{17}{4}=-\dfrac{21}{4}$.

The point is $R\left(2,\dfrac{13}{2}\right)$, so $y-\dfrac{13}{2}=-\dfrac{21}{4}(x-2)$.

Multiplying by 4 gives $4y-26=-21x+42$, hence $21x+4y-68=0$.

For $y=x^2+3x-2$, $\dfrac{dy}{dx}=2x+3$. At $x=3$, the gradient is $2(3)+3=9$.

For $x_Q=3+h$, $y_Q=(3+h)^2+3(3+h)-2=16+9h+h^2$.

$\text{gradient }PQ=\dfrac{(16+9h+h^2)-16}{(3+h)-3}=\dfrac{9h+h^2}{h}=9+h$.

As $h\to0$, point $Q$ approaches $P$, and the chord gradient $9+h$ approaches the tangent gradient $9$.

$y=\dfrac{4x^2+9}{2\sqrt{x}}=2x^{3/2}+\dfrac92x^{-1/2}$.

$\dfrac{dy}{dx}=3x^{1/2}-\dfrac94x^{-3/2}$.

$3x^{1/2}-\dfrac94x^{-3/2}=0$, so $3\sqrt{x}=\dfrac{9}{4x^{3/2}}$.

Multiply by $4x^{3/2}$: $12x^2=9$. Hence $x^2=\frac34$. Since $x>0$, $x=\frac{\sqrt3}{2}$.

For $y=2x^2+5$, $\dfrac{dy}{dx}=4x$. At $P$, $x=2$, so the tangent gradient is $8$.

If $x_Q=2+h$, then $y_Q=2(2+h)^2+5=13+8h+2h^2$.

$\text{gradient }PQ=\dfrac{(13+8h+2h^2)-13}{(2+h)-2}=\dfrac{8h+2h^2}{h}=8+2h$.

As $h$ gets closer to $0$, point $Q$ gets closer to $P$, and the chord gradient $8+2h$ tends to the tangent gradient $8$.

The normal is $y=-\dfrac1{24}x+5$. At $x=3$, $y=-\dfrac3{24}+5=\dfrac{39}{8}$.

The normal gradient is $-\dfrac1{24}$, so the tangent gradient is $24$.

$f'(3)=4(3)^2+3k+3=39+3k$.

$39+3k=24$, so $k=-5$.

With $k=-5$, $f'(x)=4x^2-5x+3$, so $f(x)=\dfrac43x^3-\dfrac52x^2+3x+c$.

$\dfrac{39}{8}=\dfrac43(27)-\dfrac52(9)+3(3)+c=\dfrac{45}{2}+c$.

$c=\dfrac{39}{8}-\dfrac{45}{2}=-\dfrac{141}{8}$.

$y=\dfrac{5x^3-8}{2x^2}=\dfrac52x-4x^{-2}$.

$\dfrac{dy}{dx}=\dfrac52+8x^{-3}=\dfrac52+\dfrac{8}{x^3}$.

At $x=2$, $\dfrac{dy}{dx}=\dfrac52+\dfrac{8}{8}=\dfrac72$.

The tangent through $P(2,4)$ is $y-4=\dfrac72(x-2)$.

$2y-8=7x-14$, so $7x-2y-6=0$.

$\sqrt[3]{8x}=2x^{1/3}$.

$\sqrt{x^3}=x^{3/2}$.

$y=\dfrac{5x^2}{2x^{1/3}}+\dfrac{x^{3/2}}{2x^{1/3}}=\dfrac52x^{5/3}+\dfrac12x^{7/6}$.

$\dfrac{dy}{dx}=\dfrac52\cdot\dfrac53x^{2/3}+\dfrac12\cdot\dfrac76x^{1/6}$.

$\dfrac{dy}{dx}=\dfrac{25}{6}x^{2/3}+\dfrac{7}{12}x^{1/6}$.

$f'(x)=ax-12x^{1/3}$, so $f''(x)=a-4x^{-2/3}$.

Since $27^{2/3}=9$, $0=a-\dfrac{4}{9}$, so $a=\dfrac49$.

With $a=\dfrac49$, $f'(x)=\dfrac49x-12x^{1/3}$. Therefore $f(x)=\dfrac29x^2-9x^{4/3}+c$.

$-8=\dfrac29-9+c$, so $c=-8+9-\dfrac29=\dfrac79$.

$f(x)=\dfrac29x^2-9x^{4/3}+\dfrac79$.

$y=\frac{x^3}{6}+4x^{1/2}-15$.

$\dfrac{dy}{dx}=\frac{3x^2}{6}+4\cdot\frac12x^{-1/2}=\frac{x^2}{2}+\frac{2}{\sqrt{x}}$.

At $x=4$, $\dfrac{dy}{dx}=\frac{16}{2}+\frac{2}{2}=9$. The normal gradient is therefore $-\frac19$.

Using $P\left(4,\frac{11}{3}\right)$: $y-\frac{11}{3}=-\frac19(x-4)$. Multiplying out gives $x+9y-37=0$.

The tangent has equation $y=5x-3\sqrt2$, so the gradient at $x=\sqrt2$ is $5$.

$f'(\sqrt2)=2\sqrt2+\dfrac8{(\sqrt2)^2}+k=2\sqrt2+4+k=5$, so $k=1-2\sqrt2$.

At $x=\sqrt2$, the tangent gives $y=5\sqrt2-3\sqrt2=2\sqrt2$.

The tangent gradient is $5$, so the normal gradient is $-\dfrac15$.

$y-2\sqrt2=-\dfrac15(x-\sqrt2)$, hence $y=-\dfrac15x+\dfrac{11\sqrt2}{5}$.

$f(x)=\int\left(2x+\dfrac8{x^2}+k\right)dx=x^2-\dfrac8x+kx+c$.

$f(x)=x^2-\dfrac8x+(1-2\sqrt2)x+c$.

$2\sqrt2=2-\dfrac8{\sqrt2}+(1-2\sqrt2)\sqrt2+c=-2-3\sqrt2+c$.

$c=2+5\sqrt2$, so $f(x)=x^2-\dfrac8x+(1-2\sqrt2)x+2+5\sqrt2$.

$y=4x^3+2x^{-1}+9$, so $\dfrac{dy}{dx}=12x^2-2x^{-2}=12x^2-\dfrac2{x^2}$.

The tangent line $y=k-5x$ has gradient $-5$, so at $P$, $12x^2-\dfrac2{x^2}=-5$.

Multiplying by $x^2$ gives $12x^4-2=-5x^2$, hence $12x^4+5x^2-2=0$.

Let $u=x^2$. Then $12u^2+5u-2=(3u+2)(4u-1)=0$.

$x^2=\dfrac14$, so $x=\dfrac12$.

$y=4\left(\dfrac12\right)^3+\dfrac{2}{1/2}+9=\dfrac12+4+9=\dfrac{27}{2}$.

Since $P$ lies on $y=k-5x$, $\dfrac{27}{2}=k-\dfrac52$, so $k=16$.

$f'(x)=6x^2-2kx+14$.

$f''(x)=12x-2k$.

At an intersection of $y=f'(x)$ and $y=f''(x)$, $f'(x)=f''(x)$. Given $x_A=5$, substitute $x=5$.

$6(5)^2-2k(5)+14=12(5)-2k$, so $164-10k=60-2k$ and $k=13$.

With $k=13$, $6x^2-26x+14=12x-26$, so $6x^2-38x+40=0$. Hence $(3x-4)(x-5)=0$.

The other point has $x=\dfrac43$. Then $y=f''(\dfrac43)=12(\dfrac43)-26=-10$.

$f'(x)=4x^{3/2}+kx^{-2}$, so $f''(x)=6x^{1/2}-2kx^{-3}$.

At $P$, $x=2$ and $f''(2)=0$, so $6\sqrt2-2k(2^{-3})=0$.

$6\sqrt2-\dfrac{k}{4}=0$, hence $k=24\sqrt2$.

$f(x)=\int\left(4x^{3/2}+24\sqrt2x^{-2}\right)dx=\dfrac85x^{5/2}-24\sqrt2x^{-1}+c$.

$P(2,8\sqrt2)$ lies on $C$, so $8\sqrt2=\dfrac85(2^{5/2})-\dfrac{24\sqrt2}{2}+c$.

$8\sqrt2=\dfrac{32\sqrt2}{5}-12\sqrt2+c$, so $c=\dfrac{68\sqrt2}{5}$.

Since $P$ has x-coordinate 4 and lies on the tangent $y=3x+4$, $P=(4,16)$.

The tangent gradient is 3, so the normal gradient is $-\dfrac13$. Hence the normal is $y-16=-\dfrac13(x-4)$.

$f''(x)=4x+x^{-1/2}$, so $f'(x)=2x^2+2\sqrt{x}+A$.

At $x=4$, $f'(4)=3$. Thus $2(16)+2(2)+A=3$, so $A=-33$.

$f(x)=\dfrac23x^3+\dfrac43x^{3/2}-33x+B$.

$16=\dfrac23(64)+\dfrac43(8)-33(4)+B=\dfrac{160}{3}-132+B$.

$B=\dfrac{284}{3}$, so $f(x)=\dfrac23x^3+\dfrac43x^{3/2}-33x+\dfrac{284}{3}$.

$y=\dfrac{x^2}{3}+4x^{-1/2}+\dfrac83x^{-1}-5$.

$\dfrac{dy}{dx}=\dfrac{2x}{3}-2x^{-3/2}-\dfrac83x^{-2}=\dfrac{2x}{3}-\dfrac{2}{x\sqrt{x}}-\dfrac{8}{3x^2}$.

At $x=4$, $\dfrac{dy}{dx}=\dfrac83-\dfrac14-\dfrac16=\dfrac94$.

The normal gradient is the negative reciprocal of $\dfrac94$, so $m_n=-\dfrac49$.

Using $P(4,3)$, $y-3=-\dfrac49(x-4)$. Multiplying by $9$ gives $9y-27=-4x+16$, hence $4x+9y-43=0$.

$f'(x)=\dfrac{32}{3x^2}+3-2\sqrt[3]{x}$. At $x=8$, $f'(8)=\dfrac{32}{3(64)}+3-2(2)=\dfrac16-1=-\dfrac56$.

Using $P(8,2)$, $y-2=-\dfrac56(x-8)$, hence $y=-\dfrac56x+\dfrac{26}{3}$.

$f'(x)=\dfrac{32}{3}x^{-2}+3-2x^{1/3}$, so $f(x)=-\dfrac{32}{3x}+3x-\dfrac32x^{4/3}+c$.

$2=-\dfrac{32}{24}+24-\dfrac32(16)+c=-\dfrac43+c$, so $c=\dfrac{10}{3}$.

$f(x)=-\dfrac{32}{3x}+3x-\dfrac32x^{4/3}+\dfrac{10}{3}$.

$y=2x^3-5x^2-\dfrac{3}{2x}+7=2x^3-5x^2-\dfrac32x^{-1}+7$.

$\dfrac{dy}{dx}=6x^2-10x+\dfrac32x^{-2}=6x^2-10x+\dfrac{3}{2x^2}$.

At $x=\dfrac12$, $y=2\left(\dfrac18\right)-5\left(\dfrac14\right)-\dfrac{3}{1}+7=3$, so $P=\left(\dfrac12,3\right)$.

At $x=\dfrac12$, $\dfrac{dy}{dx}=6\left(\dfrac14\right)-10\left(\dfrac12\right)+\dfrac{3}{2(1/4)}=\dfrac52$. The normal gradient is therefore $-\dfrac25$.

$y-3=-\dfrac25\left(x-\dfrac12\right)$. Multiplying by $5$ gives $5y-15=-2x+1$, hence $2x+5y-16=0$.

$f(x)=\dfrac{(x+5)^2}{\sqrt{x}}=\dfrac{x^2+10x+25}{x^{1/2}}=x^{3/2}+10x^{1/2}+25x^{-1/2}$.

$\int f(x)\,dx=\dfrac25x^{5/2}+\dfrac{20}{3}x^{3/2}+50x^{1/2}+c$.

$f'(x)=\dfrac32x^{1/2}+5x^{-1/2}-\dfrac{25}{2}x^{-3/2}$.

$\dfrac32x^{1/2}+5x^{-1/2}-\dfrac{25}{2}x^{-3/2}=0$.

Multiplying by $2x^{3/2}$ gives $3x^2+10x-25=0$.

$3x^2+10x-25=(3x-5)(x+5)$, so $x=\dfrac53$ or $x=-5$. Since $x>0$, $x=\dfrac53$.

$f'(x)=12x^{-1/2}+\dfrac{x}{3}-4$, so $f(x)=24\sqrt{x}+\dfrac{x^2}{6}-4x+c$.

$8=24\sqrt9+\dfrac{9^2}{6}-4(9)+c=72+\dfrac{81}{6}-36+c$.

$72+\dfrac{81}{6}-36=\dfrac{99}{2}$, so $c=8-\dfrac{99}{2}=-\dfrac{83}{2}$.

$f'(9)=\dfrac{12}{3}+\dfrac93-4=3$.

The normal gradient is the negative reciprocal of 3, so it is $-\dfrac13$.

The normal through $(9,8)$ is $y-8=-\dfrac13(x-9)$, so $y=-\dfrac13x+11$. It crosses the y-axis at $(0,11)$.

$f'(x)=2x^{-1/2}+Ax^{-2}+3$, so $f''(x)=-x^{-3/2}-2Ax^{-3}$.

$-4^{-3/2}-2A(4^{-3})=0$, so $-\dfrac18-\dfrac{A}{32}=0$ and $A=-4$.

$f'(x)=\dfrac2{\sqrt{x}}-\dfrac4{x^2}+3$.

$f(x)=4\sqrt{x}+\dfrac4x+3x+c$.

$f(12)=8\sqrt3+\dfrac13+36+c=8\sqrt3$, so $c=-\dfrac{109}{3}$.

$2x-3\sqrt{x}-5=9$, so $2x-3\sqrt{x}-14=0$.

Let $u=\sqrt{x}$, where $u>0$. Then $x=u^2$, so $2u^2-3u-14=0$.

$2u^2-3u-14=(2u-7)(u+2)$. Since $u>0$, $u=\dfrac72$, so $x=u^2=\dfrac{49}{4}$.

$f(x)=2x-3x^{1/2}-5$, so $f'(x)=2-\dfrac32x^{-1/2}$ and $f''(x)=\dfrac34x^{-3/2}$.

$\dfrac34x^{-3/2}=6$, so $x^{-3/2}=8$. Therefore $x^{3/2}=\dfrac18$.

Since $x^{3/2}=(\sqrt{x})^3$, we get $\sqrt{x}=\dfrac12$, hence $x=\dfrac14$.

$f(x)=x^3-2x^{5/2}+14$.

$f'(x)=3x^2-5x^{3/2}=x^{3/2}(3x^{1/2}-5)$. Since $x>0$, the factor $x^{3/2}$ is non-zero. Therefore $3\sqrt{x}-5=0$, so $\sqrt{x}=\dfrac53$ and $x=\dfrac{25}{9}$.

$f''(x)=6x-\dfrac{15}{2}x^{1/2}$.

$6x-\dfrac{15}{2}x^{1/2}=66$. Multiplying by 2 and dividing by 3 gives $4x-5x^{1/2}-44=0$, as required.

Let $u=\sqrt{x}$, where $u\geq0$. Then $4u^2-5u-44=0$, and $(u-4)(4u+11)=0$. Hence $u=4$ or $u=-\dfrac{11}{4}$.

Because $u=\sqrt{x}\geq0$, the value $-\dfrac{11}{4}$ is impossible. Thus $\sqrt{x}=4$ and $x=16$, which is the only solution.

$y=\dfrac23x^3-8x^2+43x-\dfrac{20}{3}$ gives $\dfrac{dy}{dx}=2x^2-16x+43$.

$2x^2-16x+43=2(x^2-8x)+43=2[(x-4)^2-16]+43$.

$\dfrac{dy}{dx}=2(x-4)^2+11$.

Since $2(x-4)^2\ge0$, the minimum value of $\dfrac{dy}{dx}$ is $11$.

The minimum occurs when $x-4=0$, so $x=4$.

At $x=4$, $y=\dfrac23(4)^3-8(4)^2+43(4)-\dfrac{20}{3}=80$.

The gradient at $S$ is $11$, so $y-80=11(x-4)$.

$y=11x+36$.

$y=2x^3+\dfrac12x^2-2x+5$, so $\dfrac{dy}{dx}=6x^2+x-2$.

At $P$, $x=0$, so the tangent gradient is $-2$.

The normal gradient is $\dfrac12$ and $P=(0,5)$, so $l:y=\dfrac12x+5$.

Set $2x^3+\dfrac12x^2-2x+5=\dfrac12x+5$.

$2x^3+\dfrac12x^2-\dfrac52x=0$. Multiplying by $2$ gives $4x^3+x^2-5x=0=x(4x^2+x-5)$.

$x(4x+5)(x-1)=0$, so the intersections have x-coordinates $0$, $-\dfrac54$ and $1$.

The point $Q$ is the left-hand intersection, so $x_Q=-\dfrac54$.

The tangent at $T$ is parallel to $l$, so its gradient is $\dfrac12$.

$6x^2+x-2=\dfrac12$, so $12x^2+2x-5=0$.

$x=\dfrac{-2\pm\sqrt{4+240}}{24}=\dfrac{-1\pm\sqrt{61}}{12}$. Since $x_T$ is positive, $x_T=\dfrac{-1+\sqrt{61}}{12}$.

At $x=4$, $f'(4)=7$. Since $f'(x)=\dfrac{2x^2+ax+b}{4\sqrt{x}}$, we get $\dfrac{32+4a+b}{8}=7$.

$32+4a+b=56$, so $4a+b=24$.

Together with $a+b=-9$, subtracting gives $3a=33$, so $a=11$ and $b=-20$.

$f'(x)=\dfrac{2x^2+11x-20}{4\sqrt{x}}=\dfrac12x^{3/2}+\dfrac{11}{4}x^{1/2}-5x^{-1/2}$.

$f(x)=\dfrac15x^{5/2}+\dfrac{11}{6}x^{3/2}-10x^{1/2}+c$.

$-5=\dfrac15(4^{5/2})+\dfrac{11}{6}(4^{3/2})-10(4^{1/2})+c=\dfrac{16}{15}+c$.

$c=-\dfrac{91}{15}$, so $f(x)=\dfrac15x^{5/2}+\dfrac{11}{6}x^{3/2}-10x^{1/2}-\dfrac{91}{15}$.

$y=f(x-3)$ translates the curve 3 units to the right, so $P(4,-5)$ becomes $Q(7,-5)$.

$3x^2-4x+20$ has discriminant $(-4)^2-4(3)(20)<0$ and positive leading coefficient, so it is always positive.

The sign of $f(x)=(x+5)(3x^2-4x+20)$ is therefore the sign of $x+5$. Hence $f(x)\ge0$ for $x\ge-5$.

$f(x)=3x^3+11x^2+100$.

$f'(x)=9x^2+22x$.

At $R(-4,84)$, $f'(-4)=9(16)+22(-4)=56$.

For the tangent at $P$ to be parallel, $f'(x)=56$.

$9x^2+22x-56=0$, so $x=\dfrac{-22\pm50}{18}$. The roots are $x=-4$ and $x=\dfrac{14}{9}$.

$x=-4$ is the point $R$, so $x_P=\dfrac{14}{9}$.

$y=f(x-3)$ translates the curve 3 units to the right, so $(-4,84)$ maps to $(-1,84)$.

$y=4f(x)$ stretches vertically by scale factor 4, so $(-4,84)$ maps to $(-4,336)$.

$(2x-1)(3x+2)=6x^2+x-2$.

$f'(x)=6x-\dfrac{6x^2+x-2}{2\sqrt{x}}=6x-3x^{3/2}-\dfrac12x^{1/2}+x^{-1/2}$.

$f'(4)=24-3(8)-\dfrac12(2)+\dfrac12=-\dfrac12$.

The normal gradient is the negative reciprocal of $-\dfrac12$, so it is $2$.

Through $P(4,12)$, $y-12=2(x-4)$, hence $y=2x+4$.

$f(x)=3x^2-\dfrac65x^{5/2}-\dfrac13x^{3/2}+2x^{1/2}+c$.

$12=3(4)^2-\dfrac65(4^{5/2})-\dfrac13(4^{3/2})+2(4^{1/2})+c$.

$12=48-\dfrac{192}{5}-\dfrac83+4+c=\dfrac{164}{15}+c$, so $c=\dfrac{16}{15}$.

$f'(x)=\dfrac{4x^2+10-7x^{1/2}}{4x^{1/2}}=x^{3/2}+\dfrac52x^{-1/2}-\dfrac74$.

At $x=4$, $f'(4)=4^{3/2}+\dfrac52(4^{-1/2})-\dfrac74=8+\dfrac54-\dfrac74=\dfrac{15}{2}$.

The normal gradient is the negative reciprocal, so $m_n=-\dfrac{2}{15}$.

Using $P(4,-1)$, $y+1=-\dfrac{2}{15}(x-4)$.

$15y+15=-2x+8$, hence $2x+15y+7=0$.

$f(x)=\int\left(x^{3/2}+\dfrac52x^{-1/2}-\dfrac74\right)dx=\dfrac25x^{5/2}+5x^{1/2}-\dfrac74x+c$.

$-1=\dfrac25(4^{5/2})+5(4^{1/2})-\dfrac74(4)+c=\dfrac{64}{5}+10-7+c$, so $c=-\dfrac{84}{5}$.

For $y=\dfrac14x^3-8x^{-1/2}$, $\dfrac{dy}{dx}=\dfrac34x^2+4x^{-3/2}$.

At $x=4$, $\dfrac{dy}{dx}=\dfrac34(16)+4(4^{-3/2})=12+\dfrac12=\dfrac{25}{2}$.

Using $P(4,12)$, $y-12=\dfrac{25}{2}(x-4)$.

Multiplying by 2 gives $2y-24=25x-100$, hence $25x-2y-76=0$.

$f'(x)=\dfrac14x^3-8x^{-1/2}$ gives $f(x)=\dfrac{x^4}{16}-16\sqrt{x}+c$.

$12=\dfrac{4^4}{16}-16\sqrt4+c=16-32+c$, so $c=28$.

$f(x)=(3x+20)(x+6)(2x-3)$ has roots $x=-\dfrac{20}{3}$, $x=-6$, and $x=\dfrac32$.

Since the cubic has positive leading coefficient, $f(x)>0$ for $-\dfrac{20}{3}<x<-6$ and for $x>\dfrac32$.

$(3x+20)(x+6)=3x^2+38x+120$.

$(3x^2+38x+120)(2x-3)=6x^3+67x^2+126x-360$.

$f'(x)=18x^2+134x+126$, so at $x=0$ the tangent gradient is $126$. Also $f(0)=-360$.

The tangent line is $y=126x-360$.

Set $6x^3+67x^2+126x-360=126x-360$. This gives $6x^3+67x^2=0$, so $x^2(6x+67)=0$.

$x=0$ is the tangent point on the y-axis, so the other intersection $P$ has $x=-\dfrac{67}{6}$.

$f(x)=2(x+1)(x-3)^2$. The curve cuts the x-axis at $(-1,0)$ and meets the x-axis at $(3,0)$ because $x=3$ is a repeated root.

$f(0)=2(1)(9)=18$, so the y-intercept is $(0,18)$.

The leading term is positive, so the cubic goes down to the left and up to the right. It crosses at $x=-1$ and touches at $x=3$.

$(x+1)(x-3)^2=(x+1)(x^2-6x+9)=x^3-5x^2+3x+9$, so $f(x)=2x^3-10x^2+6x+18$.

$f'(x)=6x^2-20x+6$.

$f'\left(\dfrac13\right)=6\left(\dfrac19\right)-20\left(\dfrac13\right)+6=0$, so the tangent is horizontal.

$f\left(\dfrac13\right)=2\left(\dfrac43\right)\left(-\dfrac83\right)^2=\dfrac{512}{27}$.

A horizontal tangent through this point has equation $y=\dfrac{512}{27}$.

$4+12x-3x^2=-3(x^2-4x)+4=-3\big((x-2)^2-4\big)+4=16-3(x-2)^2$.

The maximum occurs when $(x-2)^2=0$, so $M=(2,16)$.

Line $l_1$ passes through $O(0,0)$ and $M(2,16)$, so its gradient is $\dfrac{16}{2}=8$.

Line $l_2$ is parallel to $l_1$, so its gradient is also $8$. For $y=4+12x-3x^2$, $\dfrac{dy}{dx}=12-6x$.

Set $12-6x=8$, giving $x=\dfrac23$. Then $y=4+12\left(\dfrac23\right)-3\left(\dfrac49\right)=\dfrac{32}{3}$.

Using gradient $8$ through $\left(\dfrac23,\dfrac{32}{3}\right)$, $y-\dfrac{32}{3}=8\left(x-\dfrac23\right)$. Hence $y=8x+\dfrac{16}{3}$.

$f(x)=(2x+5)(x-3)^2$. Since $(x-3)^2\ge0$, the sign is controlled by $2x+5$, with an extra zero at $x=3$. Hence $f(x)\le0$ for $x\le-\frac52$ or $x=3$.

$(x-3)^2=x^2-6x+9$, so $f(x)=(2x+5)(x^2-6x+9)=2x^3-7x^2-12x+45$.

At the y-axis, $x=0$, so $P=(0,45)$. Also $f'(x)=6x^2-14x-12$, so the gradient at $P$ is $f'(0)=-12$.

A translation 2 units in the positive x direction means $g(x)=f(x-2)$. Therefore $g(x)=(2(x-2)+5)((x-2)-3)^2=(2x+1)(x-5)^2$.

Set $x=0$: $g(0)=(1)(25)=25$, so the y-intercept is $25$.

At $P\left(4,\frac{32}{3}\right)$, $f'(4)=5$, so the tangent has gradient $5$.

$y-\frac{32}{3}=5(x-4)$. Hence $y=5x-\frac{28}{3}$.

$f''(x)=4x^{-1/2}-3$, so $f'(x)=8x^{1/2}-3x+C$.

$5=8(2)-12+C=4+C$, so $C=1$. Therefore $f'(x)=8\sqrt{x}-3x+1$.

$f(x)=\frac{16}{3}x^{3/2}-\frac32x^2+x+D$.

Substitute $x=4$, $f(4)=\frac{32}{3}$: $\frac{32}{3}=\frac{16}{3}(8)-\frac32(16)+4+D=\frac{68}{3}+D$, so $D=-12$.

The tangent at $P$ has equation $y=10x+c$, so the gradient at $P$ is $10$.

$f'(4)=3\sqrt4+k(4)^2=6+16k=10$, so $16k=4$ and $k=\dfrac14$.

Substitute $P(4,12)$ into $y=10x+c$: $12=40+c$, so $c=-28$.

With $k=\dfrac14$, $f'(x)=3x^{1/2}+\dfrac14x^2$.

$f''(x)=\dfrac32x^{-1/2}+\dfrac12x$.

$f''(4)=\dfrac32\cdot\dfrac12+2=\dfrac34+2=\dfrac{11}{4}$.

$f(x)=\int\left(3x^{1/2}+\dfrac14x^2\right)dx=2x^{3/2}+\dfrac1{12}x^3+d$.

$12=2(4^{3/2})+\dfrac1{12}(4^3)+d=16+\dfrac{16}{3}+d$.

$d=-\dfrac{28}{3}$, so $f(x)=2x^{3/2}+\dfrac1{12}x^3-\dfrac{28}{3}$.

$f'(x)=\dfrac{(x+3)^2}{x\sqrt{x}}=\dfrac{x^2+6x+9}{x^{3/2}}=x^{1/2}+6x^{-1/2}+9x^{-3/2}$.

At $x=4$, $f'(4)=2+3+\dfrac98=\dfrac{49}{8}$.

Using $P(4,20)$, $y-20=\dfrac{49}{8}(x-4)$. Multiplying out gives $49x-8y-36=0$.

$f(x)=\int\left(x^{1/2}+6x^{-1/2}+9x^{-3/2}\right)dx=\dfrac23x^{3/2}+12x^{1/2}-18x^{-1/2}+c$.

$20=\dfrac23(8)+12(2)-\dfrac{18}{2}+c=\dfrac{61}{3}+c$, so $c=-\dfrac13$.

$f(x)=\dfrac23x\sqrt{x}+12\sqrt{x}-\dfrac{18}{\sqrt{x}}-\dfrac13$.

$f(0)=(4)(-2)(-9)=72$. Since $y=f(x)-p$ passes through $(0,50)$, $72-p=50$, so $p=22$.

If $y=f(x+q)$ passes through the origin, then $f(q)=0$. From $f(x)=(x+4)(x-2)(2x-9)$, the possible values are $q=-4,2,\dfrac92$.

$(x+4)(x-2)(2x-9)=2x^3-5x^2-34x+72$.

$f'(x)=6x^2-10x-34$.

$6x^2-10x-34<-18$, so $6x^2-10x-16<0$, or $3x^2-5x-8<0$.

$3x^2-5x-8=(3x-8)(x+1)$. Since the quadratic opens upward, it is negative between the roots: $-1<x<\dfrac83$.

$\dfrac{(3x+2)^2}{4\sqrt{x}}=\dfrac{9x^2+12x+4}{4x^{1/2}}=\dfrac94x^{3/2}+3x^{1/2}+x^{-1/2}$.

$\int\left(\dfrac94x^{3/2}+3x^{1/2}+x^{-1/2}\right)dx=\dfrac9{10}x^{5/2}+2x^{3/2}+2x^{1/2}+c$.

Factoring out $\sqrt{x}$ gives $\dfrac{\sqrt{x}(9x^2+20x+20)}{10}+c$.

Since $f'(x)=x^2+ax+b$ and the gradient at $P(3,-2)$ is $2$, $f'(3)=2$. Hence $9+3a+b=2$, so $3a+b=-7$.

$f(x)=\dfrac{x^3}{3}+\dfrac{a}{2}x^2+bx+c$. The y-intercept is $-8$, so $c=-8$.

$-2=\dfrac{27}{3}+\dfrac{9a}{2}+3b-8$, so $\dfrac{9a}{2}+3b=-3$.

From $3a+b=-7$, $b=-7-3a$. Substitute into $\dfrac{9a}{2}+3b=-3$ to get $a=-4$, and then $b=5$.

Therefore $f(x)=\dfrac{x^3}{3}-2x^2+5x-8$.

At $P$, $f'(4)=-4$, so the tangent gradient is $-4$. The normal gradient is therefore $\frac14$.

Using $P(4,-50)$: $y+50=\frac14(x-4)$, so $y=\frac14x-51$.

$f''(x)=6x^{-3/2}+x$. Therefore $f'(x)=-12x^{-1/2}+\frac12x^2+k$.

$-4=-\frac{12}{2}+\frac12(16)+k=-6+8+k=2+k$, so $k=-6$.

$f(x)=-24x^{1/2}+\frac{x^3}{6}-6x+d$.

Substitute $P(4,-50)$: $-50=-24(2)+\frac{64}{6}-24+d=-72+\frac{32}{3}+d=-\frac{184}{3}+d$. Hence $d=\frac{34}{3}$.

$f'(\sqrt3)=4(\sqrt3)^2+\dfrac6{(\sqrt3)^2}-4=12+2-4=10$, which matches the tangent gradient.

Set $4x^2+\dfrac6{x^2}-4=10$.

Multiplying by $x^2$ gives $4x^4-14x^2+6=0$, so $2x^4-7x^2+3=0$.

$2x^4-7x^2+3=(2x^2-1)(x^2-3)=0$.

$x=\sqrt3$ or $x=\dfrac1{\sqrt2}=\dfrac{\sqrt2}{2}$. The other possible x-coordinate is $\dfrac{\sqrt2}{2}$.

Given $x_P=\sqrt3$, the tangent gives $y_P=10\sqrt3-6\sqrt3=4\sqrt3$.

The normal gradient is $-\dfrac1{10}$, so $y-4\sqrt3=-\dfrac1{10}(x-\sqrt3)$.

$f(x)=\int\left(4x^2+\dfrac6{x^2}-4\right)dx=\dfrac43x^3-\dfrac6x-4x+c$.

$4\sqrt3=\dfrac43(\sqrt3)^3-\dfrac6{\sqrt3}-4\sqrt3+c=4\sqrt3-2\sqrt3-4\sqrt3+c$.

$c=6\sqrt3$, so $f(x)=\dfrac43x^3-\dfrac6x-4x+6\sqrt3$.

For $y=\dfrac27x^3+\dfrac17x^2-\dfrac52x+k$, $\dfrac{dy}{dx}=\dfrac67x^2+\dfrac27x-\dfrac52$.

At $x=-\dfrac72$, $\dfrac{dy}{dx}=\dfrac67\left(\dfrac{49}{4}\right)+\dfrac27\left(-\dfrac72\right)-\dfrac52=\dfrac{21}{2}-1-\dfrac52=7$.

Since $l$ is the normal at $A$, its gradient is the negative reciprocal of 7, so $m_l=-\dfrac17$.

At $B$, the tangent gradient is also $-\dfrac17$, so $\dfrac67x^2+\dfrac27x-\dfrac52=-\dfrac17$.

Multiplying by 14 gives $12x^2+4x-35=-2$, hence $12x^2+4x-33=0$.

$12x^2+4x-33=0$ gives $x=\dfrac{-4\pm40}{24}$, so $x=\dfrac32$ or $x=-\dfrac{11}{6}$. From the diagram, $B$ is the point to the right of the y-axis, so $x_B=\dfrac32$.

Since the y-intercept is $-1$ and the gradient is $-\dfrac17$, the line is $y=-\dfrac17x-1$.

At $x=-\dfrac72$ on the line, $y=-\dfrac17\left(-\dfrac72\right)-1=\dfrac12-1=-\dfrac12$.

$-\dfrac12=\dfrac27\left(-\dfrac72\right)^3+\dfrac17\left(-\dfrac72\right)^2-\dfrac52\left(-\dfrac72\right)+k=-\dfrac74+k$.

Therefore $k=\dfrac54$.

$\dfrac12x^2-10x+22=\dfrac12(x^2-20x)+22=\dfrac12\big((x-10)^2-100\big)+22$.

$\dfrac12\big((x-10)^2-100\big)+22=\dfrac12(x-10)^2-28$.

The minimum point is $M(10,-28)$.

The line $l$ has gradient $-\dfrac18$, so the tangent gradient at $P$ is $8$.

$\dfrac{dy}{dx}=x-10$. At $P$, $x-10=8$, so $x=18$.

When $x=18$, $y=\dfrac12(18)^2-10(18)+22=4$, so $P=(18,4)$.

Since $P$ lies on $y=k-\dfrac18x$, $4=k-\dfrac18(18)=k-\dfrac94$, so $k=\dfrac{25}{4}$.

The line is $y=\dfrac{25}{4}-\dfrac{x}{8}$.

The curve is $y=\dfrac12x^2-10x+22$.

The region lies to the right of the vertical line through $M$ and between the curve and the normal line: $10\le x\le18$ and $\dfrac12x^2-10x+22\le y\le\dfrac{25}{4}-\dfrac{x}{8}$.