Question 1
Find the second differential of the following functions:
(a)\(f(x) = 3x^6 - 2x + \dfrac{1}{x}\)
(b)\(f(x) = 5\sqrt{x}\)
(c)\(f(x) = \sqrt[3]{x^2}\)
First derivative
\(f'(x) = 18x^5 - 2 - x^{-2}\)
Answer
\(f''(x) = 90x^4 + 2x^{-3}\)
Rewrite & first derivative
\(f(x)=5x^{1/2},\quad f'(x)=\tfrac{5}{2}x^{-1/2}\)
Answer
\(f''(x) = -\dfrac{5}{4}x^{-3/2}\)
Rewrite & first derivative
\(f(x)=x^{2/3},\quad f'(x)=\tfrac{2}{3}x^{-1/3}\)
Answer
\(f''(x) = -\dfrac{2}{9}x^{-4/3}\)
Question 2
\(P = 3\pi r^3 - \dfrac{8\pi}{r^2}\). Find \(\dfrac{d^2P}{dr^2}\).
Rewrite
\(P = 3\pi r^3 - 8\pi r^{-2}\)
First derivative
\(\dfrac{dP}{dr} = 9\pi r^2 + 16\pi r^{-3}\)
Answer
\(\dfrac{d^2P}{dr^2} = 18\pi r - 48\pi r^{-4}\)
Question 3
\(f(x) = x^8\). Find \(f'(x)\), \(f''(x)\), \(f^{(3)}(x)\).
(b)Find a general equation for \(f^{(n)}(x)\), \(n \in \mathbb{Z}^+\). (HL only)
Answers
\(f'(x) = 8x^7,\quad f''(x) = 56x^6,\quad f^{(3)}(x) = 336x^5\)
Pattern
Each differentiation reduces the power by 1 and multiplies by the current power.
Answer
\(f^{(n)}(x) = \dfrac{8!}{(8-n)!}\, x^{8-n}\) for \(n \leq 8\)
Question 4
Find the equation of the tangent to the following graphs at \(x = 3\). Leave your answer in the form \(y - y_1 = m(x - x_1)\).
(a)\(f(x) = x^2 - 2\)
(b)\(f(x) = \sqrt{x}\)
(c)\(f(x) = \dfrac{2}{x}\)
Find gradient & point
\(f'(x)=2x,\quad f'(3)=6,\quad f(3)=7\)
Answer
\(y - 7 = 6(x - 3)\)
Find gradient & point
\(f'(x)=\tfrac{1}{2}x^{-1/2},\quad f'(3)=\tfrac{1}{2\sqrt{3}},\quad f(3)=\sqrt{3}\)
Answer
\(y - \sqrt{3} = \dfrac{1}{2\sqrt{3}}(x-3)\)
Find gradient & point
\(f'(x)=-2x^{-2},\quad f'(3)=-\tfrac{2}{9},\quad f(3)=\tfrac{2}{3}\)
Answer
\(y - \dfrac{2}{3} = -\dfrac{2}{9}(x-3)\)
Question 5
Find the equation of the tangent to the following graphs at \(x = -2\). Leave your answer in the form \(y = mx + c\).
(a)\(f(x) = x^3 - 3\)
(b)\(f(x) = \sqrt[3]{x^4}\)
(c)\(f(x) = \dfrac{1}{x^3}\)
Gradient & point
\(f'(x)=3x^2,\quad f'(-2)=12,\quad f(-2)=-11\)
Rewrite
\(f(x)=x^{4/3},\quad f'(x)=\tfrac{4}{3}x^{1/3},\quad f'(-2)=\tfrac{4}{3}\cdot(-2)^{1/3}=\tfrac{4\sqrt[3]{-2}}{3}\)
Point
\(f(-2)=(-2)^{4/3}=2^{4/3}=2\sqrt[3]{2}\)
Answer
\(y = \dfrac{4\sqrt[3]{-2}}{3}\,x + 2\sqrt[3]{2} + \dfrac{8\sqrt[3]{-2}}{3}\)
Gradient & point
\(f'(x)=-3x^{-4},\quad f'(-2)=-\tfrac{3}{16},\quad f(-2)=-\tfrac{1}{8}\)
Answer
\(y = -\dfrac{3}{16}x - \dfrac{7}{8}\)
Question 6
Find the equation of the normal to the following graphs at \(x = 4\). Leave your answer in the form \(y - y_1 = m(x - x_1)\).
(a)\(f(x) = x^3 - 2x\)
(b)\(f(x) = \dfrac{3}{x^2} + 5\)
(c)\(f(x) = 3x\sqrt{x}\)
Tangent gradient
\(f'(x)=3x^2-2,\quad f'(4)=46\)
Normal gradient
\(m_n = -\tfrac{1}{46},\quad f(4)=56\)
Answer
\(y - 56 = -\dfrac{1}{46}(x-4)\)
Tangent gradient
\(f'(x)=-6x^{-3},\quad f'(4)=-\tfrac{6}{64}=-\tfrac{3}{32}\)
Normal gradient
\(m_n = \tfrac{32}{3},\quad f(4)=\tfrac{3}{16}+5=\tfrac{83}{16}\)
Answer
\(y - \dfrac{83}{16} = \dfrac{32}{3}(x-4)\)
Rewrite
\(f(x)=3x^{3/2},\quad f'(x)=\tfrac{9}{2}x^{1/2},\quad f'(4)=9\)
Normal
\(m_n=-\tfrac{1}{9},\quad f(4)=24\)
Answer
\(y - 24 = -\dfrac{1}{9}(x-4)\)
Question 7
Find the equation of the normals to the following graphs at \(x = -2\). Leave your answer in the form \(ay + bx + c = 0,\ a,b,c \in \mathbb{Z}\).
(a)\(f(x) = 2x^5 + x^3\)
(b)\(f(x) = \dfrac{3}{\sqrt[4]{x}} + 1\)
(c)\(f(x) = \dfrac{5}{x^3} + x\)
Gradient & point
\(f'(x)=10x^4+3x^2,\quad f'(-2)=160+12=172,\quad f(-2)=-72\)
Normal
\(m_n=-\tfrac{1}{172}\)
Answer
\(172y + x + 72\cdot172 + 2 = 0\) i.e. \(172y + x + 12386 = 0\)
Note
\(x=-2\) gives \(x^{1/4}\) undefined in \(\mathbb{R}\). This part requires \(x>0\); use \(x=2\) instead for a valid answer.
At \(x=2\)
\(f'(x)=-\tfrac{3}{4}x^{-5/4},\quad f'(2)=-\tfrac{3}{4\cdot 2^{5/4}}\). Normal slope \(=\tfrac{4\cdot 2^{5/4}}{3}\).
Gradient & point
\(f'(x)=-15x^{-4}+1,\quad f'(-2)=-\tfrac{15}{16}+1=\tfrac{1}{16},\quad f(-2)=-\tfrac{5}{8}-2=-\tfrac{21}{8}\)
Answer
\(y + \tfrac{21}{8} = -16(x+2)\Rightarrow 8y + 128x + 277 = 0\)
Question 8
Find the equation of any horizontal tangents to the following curves:
(a)(i) \(f(x) = 3x^2 - 6\) (ii) \(f(x) = 2x^3 - 9x^2 - 24x + 5\)
(b)Use your answers to part (a)(ii) to help find the equation of any vertical normal to that curve.
Solve \(f'(x)=0\)
\(f'(x)=6x=0 \Rightarrow x=0,\quad f(0)=-6\)
Solve \(f'(x)=0\)
\(f'(x)=6x^2-18x-24=0 \Rightarrow x^2-3x-4=0 \Rightarrow (x-4)(x+1)=0\)
Points
\(f(4)=-107,\quad f(-1)=18\)
Answer
\(y = -107\) and \(y = 18\)
Vertical normal
A vertical normal occurs where the tangent is horizontal (gradient 0), so normal is vertical: \(x = c\)
Answer
\(x = 4\) and \(x = -1\)
Question 9
The graph below shows the curve \(f(x) = \sqrt[3]{x^2}\) and the tangent when \(x = 8\).
(a)Show that the tangent intersects the \(x\)-axis when \(x = -4\).
(b)Find the area of the triangle enclosed by the tangent, the \(x\)-axis, and the line \(x = 8\).
Setup
\(f(x)=x^{2/3},\quad f'(x)=\tfrac{2}{3}x^{-1/3},\quad f'(8)=\tfrac{2}{3}\cdot\tfrac{1}{2}=\tfrac{1}{3}\)
Tangent equation
\(f(8)=4\); tangent: \(y-4=\tfrac{1}{3}(x-8) \Rightarrow y=\tfrac{x}{3}+\tfrac{4}{3}\)
x-intercept
\(0=\tfrac{x}{3}+\tfrac{4}{3} \Rightarrow x=-4\) ✓
Triangle dimensions
Base along \(x\)-axis from \(x=-4\) to \(x=8\): length \(= 12\). Height \(= f(8) = 4\).
Answer
Area \(= \tfrac{1}{2} \times 12 \times 4 = 24\)