Classifying stationary points and inflection Problem set A One question at a time

Find the values of x for which:

1. (a) \(f(x)=x^3-9x^2+7x-2\) is concave-down;

   • \(f'(x)=3 x^{2} - 18 x + 7\)
   • \(f''(x)=6 x - 18\)
   • \(f''(x)<0\) gives \(6 x - 18<0\)
   • \(x<3\)

2. (b) \(f(x)=x^3+6x^2-5x+1\) is concave-up.

   • \(f'(x)=3 x^{2} + 12 x - 5\)
   • \(f''(x)=6 x + 12\)
   • \(f''(x)>0\) gives \(6 x + 12>0\)
   • \(x>-2\)

Find the coordinates of the stationary points of \(y=x^3-9x^2+15x+4\), and classify each as a local maximum or a local minimum.

• \(y'=3 x^{2} - 18 x + 15\)
• \(3x^2-18x+15=0\), \(3(x-1)(x-5)=0\), so x = 1 or x = 5.
• y-values: \(x=1\) gives \(y=11\), \(x=5\) gives \(y=-21\)
• Gradient from left to right, checked either side of each point: +, −, +
• (1, 11): gradient goes + to −, so a local maximum.
• (5, −21): gradient goes − to +, so a local minimum.

Find the coordinates of the stationary points of \(y=x^4-8x^2+3\), and classify each as a local maximum or a local minimum.

• \(y'=4 x^{3} - 16 x\)
• \(4x^3-16x=0\), \(4x(x-2)(x+2)=0\), so x = −2, 0 or 2.
• y-values: \(x=-2\) gives \(y=-13\), \(x=0\) gives \(y=3\), \(x=2\) gives \(y=-13\)
• \(y''=12 x^{2} - 16\)
• \(x=-2\): \(y''=32\) > 0, so a local minimum.
• \(x=0\): \(y''=-16\) < 0, so a local maximum.
• \(x=2\): \(y''=32\) > 0, so a local minimum.

Find the coordinates of the stationary points of \(y=8x^{1/2}-3x+2\), and classify each as a local maximum or a local minimum.

• \(y'=-3 + \frac{4}{\sqrt{x}}\)
• \(4x^{-1/2}-3=0\), so \(x^{1/2}=\frac43\) and \(x=\frac{16}{9}\).
• y-values: \(x=\frac{16}{9}\) gives \(y=\frac{22}{3}\) = 7.33
• \(y''=- \frac{2}{x^{\frac{3}{2}}}\)
• \(x=\frac{16}{9}\): \(y''=- \frac{27}{32}\) = −0.844 < 0, so a local maximum.

Find the coordinates of the stationary points of \(y=e^x-4x\), and classify each as a local maximum or a local minimum. Give coordinates to 3 significant figures.

• \(y'=e^{x} - 4\)
• \(e^x-4=0\), so \(e^x=4\) and \(x=\ln 4\).
• y-values: \(x=\ln{\left(4 \right)}\) gives \(y=4 - 8 \ln{\left(2 \right)}\) = −1.55
• \(y''=e^{x}\)
• \(x=\ln{\left(4 \right)}\): \(y''=4\) > 0, so a local minimum.

Find the coordinates of the stationary points of \(y=5\ln x-2x\), and classify each as a local maximum or a local minimum. Give coordinates to 3 significant figures.

• \(y'=-2 + \frac{5}{x}\)
• \(\frac5x-2=0\), so \(x=\frac52\).
• y-values: \(x=\frac{5}{2}\) gives \(y=-5 + 5 \ln{\left(\frac{5}{2} \right)}\) = −0.419
• \(y''=- \frac{5}{x^{2}}\)
• \(x=\frac{5}{2}\): \(y''=- \frac{4}{5}\) = −0.8 < 0, so a local maximum.

\(f(x)=2x^3+kx^2-5x+4\) and \(f''(3)=20\). Find the value of k.

• \(f'(x)=2 k x + 6 x^{2} - 5\)
• \(f''(x)=2 k + 12 x\)
• \(f''(3)=2 k + 36\) and this equals 20
• \(2 k + 36=20\), so \(k=-8\)

\(f(x)=x^3+ax^2+bx+7\), with \(f''(-2)=-18\) and \(f'(3)=14\). Find the values of a and b.

• \(f'(x)=2 a x + b + 3 x^{2}\), \(f''(x)=2 a + 6 x\)
• \(f''(-2)=-18\): \(2 a - 12=-18\), so \(a=-3\)
• \(f'(3)=14\): \(6 a + b + 27=14\)
• Substitute \(a=-3\): \(b=5\)

\(y=xe^{3x}\).

1. (a) Find \(d^2y/dx^2\).

   • \(u=x,\ v=e^{3x}\): \(u'=1,\ v'=3e^{3x}\)
   • \(y'=\left(3 x + 1\right) e^{3 x}\)
   • \(y''=3 \left(3 x + 2\right) e^{3 x}\)

2. (b) Find the value of \(d^2y/dx^2\) when \(x=0\).

   • \(y''(0)=6\)

A company makes x hundred phones a week. Its weekly profit is \(P=-x^3+9x^2-15x\) thousand dollars, for \(0\le x\le 7\). Find the value of x that gives the maximum, using the second derivative to show that it is a maximum, and state the maximum value.

• \(P'=- 3 x^{2} + 18 x - 15\)
• \(P'=0\) gives x = 1 or x = 5.
• \(P''=18 - 6 x\)
• \(x=1\): \(P''=12\) > 0, a minimum. \(x=5\): \(P''=-12\) < 0, a maximum.
• Maximum: x = 5, P = 25 thousand dollars.

The total sales of a new game, S thousand copies, t months after its launch are modelled by \(S=-t^3+12t^2+10t\) for \(0\le t\le 12\).

1. (a) Find \(S''\).

   • \(S'=- 3 t^{2} + 24 t + 10\)
   • \(S''=24 - 6 t\)

2. (b) Find the value of t at which \(S''=0\), and show that the concavity changes there.

   • \(S''=0\): \(24 - 6 t=0\), so \(t=4\)
   • \(S''\) is positive for \(t<4\) and negative for \(t>4\): the concavity changes, so this is a point of inflection.

3. (c) Explain why the sales are growing fastest at this time, and find that rate of growth.

   • \(S'\) is the rate of growth. It is greatest where the gradient stops increasing and starts decreasing: at the inflection.
   • \(S'(4)=58\) thousand copies per month.

Sketch a curve for \(x>0\) where \(f(x)>0\), \(f'(x)>0\) and \(f''(x)<0\).

• f(x) > 0: the curve is above the x-axis.
• f′(x) > 0: it is increasing, rising from left to right.
• f″(x) < 0: the gradient is decreasing, so it is concave-down: it rises more and more slowly.
• One example is y = √x for x > 0.

Find the stationary points of \(y=3x^4+4x^3\). Classify each one. If the second derivative test does not decide, say what you do instead.

• \(y'=12 x^{3} + 12 x^{2}=12 x^{2} \left(x + 1\right)\), so x = −1 or x = 0.
• \(y''=36 x^{2} + 24 x\)
• \(y''(-1)=12\) so a local minimum at \((-1,-1)\).
• \(y''(0)=0\): the test fails. Check the gradient: \(y'(-1/2)=3/2\) and \(y'(1/2)=9/2\) have the same sign, so the gradient does not change sign.