Question 1
A graph of a function \(f(x)\) has stationary points at \(x = -1\) (local max, \(f(-1) = 4\)) and \(x = 3\) (local min, \(f(3) = -2\)), and passes through \((0, 2)\).
(a)For the graph \(f(x)\), state the interval(s) where \(f'(x) < 0\).
(b)State the values of \(x\) where \(f'(x) = 0\).
Answer
\(-1 < x < 3\) (between the local max and local min)
Answer
\(x = -1\) and \(x = 3\)
Question 2
A function \(f(x)\) is defined below: \[f(x) = \dfrac{8}{x^2},\quad x \neq 0\]
(a)Write down the equations of the vertical and horizontal asymptotes.
(b)Show using calculus that \(f(x)\) is a decreasing function for all \(x > 0\).
Answer
Vertical: \(x = 0\); Horizontal: \(y = 0\)
Differentiate
\(f'(x) = -16x^{-3} = -\dfrac{16}{x^3}\)
Argument
For \(x > 0\): \(x^3 > 0\), so \(f'(x) = -\dfrac{16}{x^3} < 0\) for all \(x > 0\). Hence \(f\) is decreasing. ✓
Question 3
A function \(f(x) = 3bx^2 + 2x\) where \(b\) is a positive constant. For what values of \(x\) is \(f(x)\) a decreasing function?
Differentiate
\(f'(x) = 6bx + 2\)
Set \(f'(x) < 0\)
\(6bx + 2 < 0 \Rightarrow x < -\dfrac{1}{3b}\)
Answer
\(x < -\dfrac{1}{3b}\)
Question 4
Use the first derivative test to classify any stationary points of the following graphs:
(a)\(f(x) = 2x^2 - 8x + 3\)
(b)\(f(x) = -x^3 + 3x^2 + 9x - 5\)
(c)\(f(x) = \dfrac{x^3}{3} - x^2 - 8x + 2\)
(d)\(f(x) = \dfrac{x^3}{3} - x^2 - 3x + 1\)
Differentiate & solve
\(f'(x) = 4x-8 = 0 \Rightarrow x=2,\quad f(2)=-5\)
Sign test
\(x<2: f'<0;\quad x>2: f'>0\) → minimum
Answer
Local minimum at \((2,\,-5)\)
Differentiate & solve
\(f'(x) = -3x^2+6x+9 = -3(x^2-2x-3) = -3(x-3)(x+1) = 0\)
\(\Rightarrow x = 3\) or \(x = -1\)
Sign tests
At \(x=-1\): \(f'(-2)=-3(4+4-3)<0\to f'(0)=9>0\) → local min at \((-1,-10)\)
At \(x=3\): \(f'(1)=12>0\to f'(4)=-15<0\) → local max at \((3,22)\)
Answer
Local min at \((-1,\,-10)\); local max at \((3,\,22)\)
Differentiate & solve
\(f'(x) = x^2-2x-8 = (x-4)(x+2) = 0 \Rightarrow x=4\) or \(x=-2\)
Answer
Local max at \(\!\left(-2,\,\tfrac{34}{3}\right)\); local min at \(\left(4,\,-\tfrac{46}{3}\right)\)
Differentiate & solve
\(f'(x)=x^2-2x-3=(x-3)(x+1)=0 \Rightarrow x=3\) or \(x=-1\)
Answer
Local max at \(\left(-1,\,\tfrac{8}{3}\right)\); local min at \(\left(3,\,-8\right)\)
Question 5
Use the second derivative test to classify any stationary points of the following graphs:
(a)\(f(x) = 2x^3 - 9x^2\)
(b)\(f(x) = \dfrac{x^4}{4} - 2x^2 + 1\)
(c)\(f(x) = 2x^3 + 3x^2 - 12x\)
(d)\(x^3 - 3x^2 - 9x + 5\)
Solve \(f'=0\)
\(f'(x)=6x^2-18x=6x(x-3)=0 \Rightarrow x=0,\;x=3\)
Second derivative
\(f''(x)=12x-18\). \(f''(0)=-18<0\) → max; \(f''(3)=18>0\) → min
Answer
Local max at \((0,0)\); local min at \((3,-27)\)
Solve \(f'=0\)
\(f'(x)=x^3-4x=x(x^2-4)=x(x-2)(x+2)=0 \Rightarrow x=0,\pm2\)
Second derivative
\(f''(x)=3x^2-4\). \(f''(0)=-4<0\) → max; \(f''(\pm2)=8>0\) → min
Answer
Local max at \((0,1)\); local min at \((\pm2,-3)\)
Solve \(f'=0\)
\(f'(x)=6x^2+6x-12=6(x+2)(x-1)=0 \Rightarrow x=-2,\;x=1\)
Second derivative
\(f''(x)=12x+6\). \(f''(-2)=-18<0\) → max; \(f''(1)=18>0\) → min
Answer
Local max at \((-2,20)\); local min at \((1,-7)\)
Solve \(f'=0\)
\(f'(x)=3x^2-6x-9=3(x-3)(x+1)=0 \Rightarrow x=3,\;x=-1\)
Second derivative
\(f''(x)=6x-6\). \(f''(-1)=-12<0\) → max; \(f''(3)=12>0\) → min
Answer
Local max at \((-1,10)\); local min at \((3,-22)\)
Question 6
Classify any stationary points of the following graph: \[y = \dfrac{x^2}{x - 1},\quad x \neq 1\]
Quotient rule
\(\dfrac{dy}{dx} = \dfrac{2x(x-1) - x^2}{(x-1)^2} = \dfrac{x^2 - 2x}{(x-1)^2} = \dfrac{x(x-2)}{(x-1)^2}\)
Solve \(=0\)
\(x=0\) or \(x=2\) (denominator \(\neq 0\) at these points)
Sign analysis
\(x=0\): \(f'(-1)=\tfrac{3}{4}>0\to f'(\tfrac{1}{2})=-1<0\) → local max
\(x=2\): \(f'(\tfrac{3}{2})=-3<0\to f'(3)=\tfrac{3}{4}>0\) → local min
Answer
Local max at \((0,0)\); local min at \((2,4)\)
Question 7
The graph \(f(x) = x^3 + ax^2 + bx + 6\) has a stationary point at \((2, -2)\).
(a)Find \(a\) and \(b\).
(b)Classify all stationary points.
Two equations
From \(f(2)=-2\): \(8+4a+2b+6=-2 \Rightarrow 4a+2b=-16 \Rightarrow 2a+b=-8 \quad\cdots(1)\)
From \(f'(2)=0\): \(f'(x)=3x^2+2ax+b\Rightarrow 12+4a+b=0 \Rightarrow 4a+b=-12 \quad\cdots(2)\)
Solve simultaneously
\((2)-(1): 2a=-4\Rightarrow a=-2,\quad b=-8-2(-2)=-4\)
Answer
\(a = -2,\quad b = -4\)
Find stationary points
\(f'(x)=3x^2-4x-4=(3x+2)(x-2)=0 \Rightarrow x=-\tfrac{2}{3}\) or \(x=2\)
Second derivative
\(f''(x)=6x-4\). \(f''\!\left(-\tfrac{2}{3}\right)=-8<0\) → max; \(f''(2)=8>0\) → min
Answer
Local max at \(\left(-\tfrac{2}{3},\,f\!\left(-\tfrac{2}{3}\right)\right)\); confirmed local min at \((2,-2)\)