Mixed practice: stationary points, concavity, parameters and models Problem set B One question at a time

Sketch a graph of \(y=f(x)\) where \(f'(x)>0\) for all x, \(f''(x)>0\) for \(x<-1\) and \(f''(x)<0\) for \(x>-1\). Mark the point where the bend changes.

• \(f'>0\): the curve goes up from left to right everywhere.
• \(f''>0\) for \(x<-1\): the gradient is rising, so the curve gets steeper. It bends up.
• \(f''<0\) for \(x>-1\): the gradient is falling (but still positive), so the curve flattens. It bends down.
• \(f''\) changes sign at \(x=-1\): mark a point of inflection there. The gradient is at its greatest at that point.

The curve \(y=2x^3+px^2+5x-4\) is concave up for \(x>3\) only.

1. (a) Find the value of p.

   • \(y''=2 \left(p + 6 x\right)\)
   • \(y''=0\) at \(x=3\): \(2 p + 36=0\)
   • \(p=-18\)

2. (b) What can you say about the curve at \(x=3\)?

   • \(y''=12 x - 36\), which is 0 at \(x=3\) and changes sign there.
   • The curve has a point of inflection at \(x=3\).

The function \(f(x)=x^n\) has \(f''(1)=30\). Find the possible values of n.

• \(f'=nx^{n-1}\)
• \(f''=n(n-1)x^{n-2}\)
• \(f''(1)=n(n-1)=30\), so \(n^2-n-30=0\)
• \((n-6)(n+5)=0\)
• \(n=6\) or \(n=-5\)

A curve has a minimum point. Find the values of b and c in each case.

1. (a) \(y=x^2+bx+c\) has a minimum at \((4,1)\).

   • \(y'=2x+b=0\) at \(x=4\): \(8+b=0\), so \(b=-8\)
   • \(y=1\) at \(x=4\): \(16+(-8)(4)+c=1\)
   • \(c=17\)

2. (b) \(y=x^4+bx+c\) has a minimum at \((1,3)\).

   • \(y'=4x^3+b=0\) at \(x=1\): \(4+b=0\), so \(b=-4\)
   • \(y=3\) at \(x=1\): \(1+(-4)(1)+c=3\)
   • \(c=6\)

Find the stationary points of each curve and classify them.

1. (a) \(y=3x-8\sqrt x\), for \(x>0\). Give exact values.

   • \(y'=3 - \frac{4}{\sqrt{x}}=0\), so \(\sqrt x=\frac{4}{3}\) and \(x=\frac{16}{9}\)
   • \(y''=\frac{2}{x^{\frac{3}{2}}}>0\) for all \(x>0\), so it is a minimum
   • \(y=- \frac{16}{3}\)

2. (b) \(y=\dfrac{x}{x^2+9}\). Give exact values.

   • \(y'=\dfrac{9-x^2}{(x^2+9)^2}=0\) when \(x^2=9\), so \(x=\pm3\)
   • \(y''=\frac{2 x \left(x^{2} - 27\right)}{\left(x^{2} + 9\right)^{3}}\): \(y''(-3)=\frac{1}{54}>0\) and \(y''(3)=- \frac{1}{54}<0\)
   • Minimum at \(x=-3\), maximum at \(x=3\)

3. (c) \(y=x+2\sin x\), for \(0<x<2\pi\). Give exact values.

   • \(y'=2 \cos{\left(x \right)} + 1=0\), so \(x\) is \(\frac{2 \pi}{3}\) or \(\frac{4 \pi}{3}\)
   • \(y''=- 2 \sin{\left(x \right)}\): \(y''(\frac{2 \pi}{3})=- \sqrt{3}<0\) and \(y''(\frac{4 \pi}{3})=\sqrt{3}>0\)
   • Maximum at \(x=\frac{2 \pi}{3}\), minimum at \(x=\frac{4 \pi}{3}\)

Show each of these results.

1. (a) The graph of \(y=4x^2+bx+c\) is concave up for every value of x.

   • \(y'=8x+b\)
   • \(y''=8\)
   • \(8>0\) for all x, so the graph is always concave up.

2. (b) If \(y=4\sin 5x+\cos 5x\), then \(\dfrac{d^2y}{dx^2}=-25y\).

   • \(y'=- 5 \sin{\left(5 x \right)} + 20 \cos{\left(5 x \right)}\)
   • \(y''=- 25 \left(4 \sin{\left(5 x \right)} + \cos{\left(5 x \right)}\right)\)
   • \(=-25\left(4\sin 5x+\cos 5x\right)=-25y\)

3. (c) The graph of \(y=\ln(a+3x)\) is always concave down.

   • \(y'=\dfrac{3}{a+3x}=3(a+3x)^{-1}\)
   • \(y''=-3\cdot3(a+3x)^{-2}=\dfrac{-9}{(a+3x)^2}\)
   • The denominator is a square, so it is positive. Then \(y''<0\) wherever the graph exists: always concave down.

The curve is \(y=x^3+6x^2-4x+1\).

1. (a) Find the interval in which the gradient is decreasing.

   • \(y'=3 x^{2} + 12 x - 4\)
   • \(y''=6 \left(x + 2\right)\)
   • The gradient is decreasing when \(y''<0\): \(6 \left(x + 2\right)<0\), so \(x<-2\)

2. (b) Hence find the coordinates of the point of inflection.

   • \(y''=0\) at \(x=-2\), and \(y''\) changes from negative to positive there
   • \(y(-2)=25\)
   • Point of inflection \((-2,25)\)

The curve is \(y=x^5\ln x\), for \(x>0\).

1. (a) Find the x-coordinate of the point where \(y''=3\). Use the calculator to solve the equation.

   • \(y'=5x^4\ln x+x^{4}\)
   • \(y''=x^{3} \left(20 L + 9\right)\) with \(L=\ln x\), so \(y''=x^3(20\ln x+9)\)
   • From the calculator: \(x^{3}(20\ln x+9)=3\) gives \(x=0.829\)

2. (b) The curve has a point of inflection. Find its x-coordinate.

   • \(y''=0\): \(20\ln x+9=0\) (as \(x>0\))
   • \(\ln x=-\dfrac{9}{20}\), so \(x=e^{-9/20}=0.638\)
   • y″ changes sign here, so it is a point of inflection.

The function f is defined by \(f(x)=\dfrac{\ln x}{x^4}\), for \(x>0\).

1. (a) Find \(f'(x)\).

   • Quotient rule: \(f'=\dfrac{x^4\cdot\frac1x-4x^3\ln x}{x^{8}}=\dfrac{x^{3}(1-4\ln x)}{x^{8}}\)
   • \(f'(x)=\dfrac{1-4\ln x}{x^5}\)

2. (b) Find the exact coordinates of the stationary point.

   • \(f'=0\): \(1-4\ln x=0\), so \(\ln x=\dfrac14\) and \(x=e^{1/4}\)
   • \(x^4=e\), so \(f=\dfrac{1/4}{e}=\dfrac1{4e}\)

3. (c) Show that \(f''(x)=\dfrac{20\ln x-9}{x^6}\).

   • Quotient rule: \(f''=\dfrac{-\frac4x\cdot x^5-5x^4(1-4\ln x)}{x^{10}}\)
   • \(=\dfrac{-4x^4-5x^4+20x^4\ln x}{x^{10}}\)
   • \(=\dfrac{20\ln x-9}{x^6}\)

4. (d) Hence classify the stationary point.

   • At \(x=e^{1/4}\), \(\ln x=\dfrac14\), so the numerator is \(-4<0\)
   • The denominator is positive, so \(f''<0\): a local maximum.

A factory makes x hundred phone cases a week, where \(x>0\). The weekly profit, in thousands of dollars, is modelled by \(P=-x^3+15x^2-48x-10\).

1. (a) Find the value of x that gives the greatest profit. Show that it is a maximum.

   • \(P'=- 3 x^{2} + 30 x - 48=0\): \(- 3 \left(x - 8\right) \left(x - 2\right)=0\)
   • \(x=2\) or \(x=8\)
   • \(P''=6 \left(5 - x\right)\): \(P''(8)=-18<0\), so \(x=8\) is a maximum

2. (b) Find the greatest profit.

   • \(P(8)=54\) thousand dollars

A paddling pool is filled from a hose whose pressure changes. The volume of water, V litres, after t minutes is modelled by \(V=t^3-9t^2+30t\), for \(t\ge0\). Find the time at which the pool is filling most slowly, and state the minimum rate of filling.

• The rate of filling is \(V'=3 t^{2} - 18 t + 30\)
• Minimise \(V'\): \(V''=6 \left(t - 3\right)=0\) gives \(t=3\)
• \(V'''=6>0\), so \(V'\) has a minimum at \(t=3\)
• \(V'(3)=3\) litres per minute

The number of subscribers to a video channel, P thousand, is modelled by \(P=\dfrac{20t^2}{16+t^2}\), where t is the time in months since the channel started.

1. (a) Find \(\dfrac{dP}{dt}\) and use it to find the time at which the number is increasing most quickly. Give your answer to 3 significant figures.

   • Quotient rule: \(\dfrac{dP}{dt}=\dfrac{640t}{(16+t^2)^2}\)
   • Greatest rate: \(P''=- \frac{640 \left(3 t^{2} - 16\right)}{\left(t^{2} + 16\right)^{3}}=0\)
   • \(16-3t^2=0\), so \(t=\sqrt{16/3}=2.31\)
   • \(P''\) changes from positive to negative here, so \(P'\) is a maximum.

2. (b) Find the number at this time and the greatest rate of increase.

   • \(P=20t^2/(16+t^2)\) at \(t^2=16/3\): \(P=5.00\) thousand
   • Rate: \(P'=3.25\) thousand per month

Find the stationary points of the curve \(y=px+\dfrac{9}{x}\), where \(p>0\), and classify them. Give your answers in terms of p.

• \(y'=p-\dfrac{9}{x^2}=0\), so \(x^2=\dfrac{9}{p}\) and \(x=\pm\dfrac{3}{\sqrt p}\)
• \(y=\pm\left(\sqrt p\cdot3+3\sqrt p\right)=\pm6\sqrt p\)
• \(y''=\dfrac{18}{x^3}\): positive for \(x>0\), negative for \(x<0\)
• \(x=\dfrac{3}{\sqrt p}\) is a minimum; \(x=-\dfrac{3}{\sqrt p}\) is a maximum

A cuboid has a square base of side x cm and height \((a-3x)\) cm, where a is a positive constant and \(0<x<\dfrac{a}{3}\).

1. (a) Find the value of x, in terms of a, that gives the greatest volume. Justify that it is a maximum.

   • \(V=x^2(a-3x)=a x^{2} - 3 x^{3}\)
   • \(V'=2 a x - 9 x^{2}=- x \left(- 2 a + 9 x\right)=0\), so \(x=\frac{2 a}{9}\) (\(x=0\) is not allowed)
   • \(V''=2 \left(a - 9 x\right)\), and \(V''\left(\frac{2 a}{9}\right)=- 2 a<0\) because \(a>0\)
   • So the volume is a maximum.

2. (b) Find the greatest volume in terms of a.

   • \(V\left(\frac{2 a}{9}\right)=\frac{4 a^{3}}{243}\)

Answer both parts.

1. (a) Find the range of \(f(x)=e^{3x}-ax\), where \(a>0\), in terms of a.

   • \(f'=3e^{3x}-a=0\), so \(e^{3x}=\dfrac{a}{3}\) and \(x=\dfrac13\ln\dfrac{a}{3}\)
   • \(f''=9e^{3x}>0\), so this is the minimum
   • \(f_{\min}=\dfrac{a}{3}-\dfrac{a}{3}\ln\dfrac{a}{3}\)
   • Range: \(f(x)\ge\dfrac{a}{3}\left(1-\ln\dfrac{a}{3}\right)\)

2. (b) A model for the potential energy U joules of a particle at position x metres is \(U=e^{3x}+e^{-x}\). Find the exact minimum value of U.

   • \(U'=3e^{3x}-e^{-x}=0\), so \(e^{4x}=\dfrac{1}{3}\) and \(x=\dfrac{1}{4}\ln\dfrac{1}{3}\)
   • \(e^{3x}=\left(\dfrac{1}{3}\right)^{3/4}\) and \(e^{-x}=\left(\dfrac{1}{3}\right)^{-1/4}=3\left(\dfrac{1}{3}\right)^{3/4}\)
   • \(U''=9e^{3x}+e^{-x}>0\): a minimum
   • \(U_{\min}=\left(1+3\right)\left(\dfrac{1}{3}\right)^{3/4}=4\left(\dfrac{1}{3}\right)^{3/4}=1.75\) J

Answer both parts.

1. (a) The curve \(y=x^3+4x^2+bx-5\) has a positive gradient at every point (it is strictly increasing and never flat). Find the condition on b.

   • \(y'=b + 3 x^{2} + 8 x\)
   • A positive gradient at every point means \(y'>0\) for all x, so \(y'=0\) must have no real roots: \(\Delta<0\)
   • \(\Delta=(8)^2-4(3)(b)=64 - 12 b<0\)
   • \(b>\frac{16}{3}\)

2. (b) The curve \(y=ax^3-bx^2\), where a and b are positive, is increasing for \(x<0\) and for \(x>9\). Evaluate \(\dfrac{b}{a}\).

   • \(y'=3ax^2-2bx=x(3ax-2b)\)
   • \(y'\) changes sign at \(x=0\) and at \(x=\dfrac{2b}{3a}\), which is the \(x=9\) boundary
   • \(\dfrac{2b}{3a}=9\), so \(\dfrac{b}{a}=\frac{27}{2}\)

A curve has \(y''=12x-6\). Its gradient at \(x=0\) is 3 and it passes through the point \((1,-1)\). Find y in terms of x.

• \(y'=\int(12x-6)\,dx=6 x^{2} - 6 x+C\)
• \(y'(0)=3\) gives \(C=3\), so \(y'=6 x^{2} - 6 x + 3\)
• \(y=2 x^{3} - 3 x^{2} + 3 x+D\)
• \(y(1)=-1\) gives \(D=-3\)
• \(y=2 x^{3} - 3 x^{2} + 3 x - 3\)

The graph shows \(y=f'(x)\). It crosses the x-axis at \(x=3\) and touches the x-axis at \(x=-1\).

1. (a) Write down the x-coordinate of each local minimum of \(f\), and of each local maximum of \(f\).

   • Local extrema of f are where \(f'\) crosses the x-axis.
   • At \(x=3\), \(f'\) goes from negative to positive: local minimum.
   • At \(x=-1\), \(f'\) touches the axis and stays negative: not a maximum or minimum.

2. (b) Write down the x-coordinate of each point of inflection of \(f\).

   • Inflections of f are where \(f'\) has a local maximum or minimum (f″ changes sign).
   • \(f'\) has a local maximum at \(x=-1\) and a local minimum at \(x=\dfrac53\).

3. (c) Which of these points of inflection are stationary points of inflection? Explain.

   • At \(x=-1\), \(f'=0\) as well as \(f''=0\): a stationary point of inflection.
   • At \(x=\dfrac53\), \(f'=-\dfrac{256}{27}\), not 0: not stationary.