Classifying stationary points and inflection Problem set B Worked solutions

Find the values of x for which:

1. (a) \(f(x)=x^3-12x^2+4x+3\) is concave-down;

   • \(f'(x)=3 x^{2} - 24 x + 4\)
   • \(f''(x)=6 x - 24\)
   • \(f''(x)<0\) gives \(6 x - 24<0\)
   • \(x<4\)

2. (b) \(f(x)=x^3+15x^2-2x+8\) is concave-up.

   • \(f'(x)=3 x^{2} + 30 x - 2\)
   • \(f''(x)=6 x + 30\)
   • \(f''(x)>0\) gives \(6 x + 30>0\)
   • \(x>-5\)

Find the coordinates of the stationary points of \(y=x^3+6x^2-15x+1\), and classify each as a local maximum or a local minimum.

• \(y'=3 x^{2} + 12 x - 15\)
• \(3x^2+12x-15=0\), \(3(x+5)(x-1)=0\), so x = −5 or x = 1.
• y-values: \(x=-5\) gives \(y=101\), \(x=1\) gives \(y=-7\)
• Gradient from left to right, checked either side of each point: +, −, +
• (−5, 101): gradient goes + to −, so a local maximum.
• (1, −7): gradient goes − to +, so a local minimum.

Find the coordinates of the stationary points of \(y=x^4-32x^2-7\), and classify each as a local maximum or a local minimum.

• \(y'=4 x^{3} - 64 x\)
• \(4x^3-64x=0\), \(4x(x-4)(x+4)=0\), so x = −4, 0 or 4.
• y-values: \(x=-4\) gives \(y=-263\), \(x=0\) gives \(y=-7\), \(x=4\) gives \(y=-263\)
• \(y''=12 x^{2} - 64\)
• \(x=-4\): \(y''=128\) > 0, so a local minimum.
• \(x=0\): \(y''=-64\) < 0, so a local maximum.
• \(x=4\): \(y''=128\) > 0, so a local minimum.

Find the coordinates of the stationary points of \(y=10x^{1/2}-4x+1\), and classify each as a local maximum or a local minimum.

• \(y'=-4 + \frac{5}{\sqrt{x}}\)
• \(5x^{-1/2}-4=0\), so \(x^{1/2}=\frac54\) and \(x=\frac{25}{16}\).
• y-values: \(x=\frac{25}{16}\) gives \(y=\frac{29}{4}\) = 7.25
• \(y''=- \frac{5}{2 x^{\frac{3}{2}}}\)
• \(x=\frac{25}{16}\): \(y''=- \frac{32}{25}\) = −1.28 < 0, so a local maximum.

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

• \(y'=5 - e^{x}\)
• \(5-e^x=0\), so \(e^x=5\) and \(x=\ln 5\).
• y-values: \(x=\ln{\left(5 \right)}\) gives \(y=-5 + 5 \ln{\left(5 \right)}\) = 3.05
• \(y''=- e^{x}\)
• \(x=\ln{\left(5 \right)}\): \(y''=-5\) < 0, so a local maximum.

Find the coordinates of the stationary points of \(y=4x-6\ln x\), and classify each as a local maximum or a local minimum. Give coordinates to 3 significant figures.

• \(y'=4 - \frac{6}{x}\)
• \(4-\frac6x=0\), so \(x=\frac32\).
• y-values: \(x=\frac{3}{2}\) gives \(y=6 - 6 \ln{\left(\frac{3}{2} \right)}\) = 3.57
• \(y''=\frac{6}{x^{2}}\)
• \(x=\frac{3}{2}\): \(y''=\frac{8}{3}\) = 2.67 > 0, so a local minimum.

\(f(x)=x^3+kx^2+2x-1\) and \(f''(-2)=-4\). Find the value of k.

• \(f'(x)=2 k x + 3 x^{2} + 2\)
• \(f''(x)=2 k + 6 x\)
• \(f''(-2)=2 k - 12\) and this equals −4
• \(2 k - 12=-4\), so \(k=4\)

\(f(x)=x^3+ax^2+bx-2\), with \(f''(-3)=-14\) and \(f'(2)=13\). Find the values of a and b.

• \(f'(x)=2 a x + b + 3 x^{2}\), \(f''(x)=2 a + 6 x\)
• \(f''(-3)=-14\): \(2 a - 18=-14\), so \(a=2\)
• \(f'(2)=13\): \(4 a + b + 12=13\)
• Substitute \(a=2\): \(b=-7\)

\(y=xe^{-2x}\).

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

   • \(u=x,\ v=e^{-2x}\): \(u'=1,\ v'=-2e^{-2x}\)
   • \(y'=- \left(2 x - 1\right) e^{- 2 x}\)
   • \(y''=4 \left(x - 1\right) e^{- 2 x}\)

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

   • \(y''(0)=-4\)

A farmer puts x kg of fertiliser on each hectare. The crop yield is \(Y=-x^3+12x^2-36x+50\) tonnes, for \(0< x\le 8\). 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.

• \(Y'=- 3 x^{2} + 24 x - 36\)
• \(Y'=0\) gives x = 2 or x = 6.
• \(Y''=24 - 6 x\)
• \(x=2\): \(Y''=12\) > 0, a minimum. \(x=6\): \(Y''=-12\) < 0, a maximum.
• Maximum: x = 6, Y = 50 tonnes.

The number of students who have heard a rumour, R hundred, t days after it starts is modelled by \(R=-t^3+18t^2+5t\) for \(0\le t\le 18\).

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

   • \(R'=- 3 t^{2} + 36 t + 5\)
   • \(R''=36 - 6 t\)

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

   • \(R''=0\): \(36 - 6 t=0\), so \(t=6\)
   • \(R''\) is positive for \(t<6\) and negative for \(t>6\): the concavity changes, so this is a point of inflection.

3. (c) Explain why the rumour is spreading fastest at this time, and find that rate of growth.

   • \(R'\) is the rate of growth. It is greatest where the gradient stops increasing and starts decreasing: at the inflection.
   • \(R'(6)=113\) hundred students per day.

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 decreasing, falling from left to right.
• f″(x) > 0: the gradient is increasing towards 0, so it is concave-up: it falls more and more slowly.
• One example is y = 1/x for x > 0.

Find the stationary points of \(y=3x^4-8x^3\). Classify each one. If the second derivative test does not decide, say what you do instead.

• \(y'=12 x^{3} - 24 x^{2}=12 x^{2} \left(x - 2\right)\), so x = 0 or x = 2.
• \(y''=36 x^{2} - 48 x\)
• \(y''(0)=0\): the test fails. Check the gradient: \(y'(-1/2)=-15/2\) and \(y'(1/2)=-9/2\) have the same sign, so the gradient does not change sign.
• \(y''(2)=48\) so a local minimum at \((2,-16)\).