The second derivative and concavity Problem set A Worked solutions

Find \(f''(2)\) for each function.

1. (a) \(f(x)=7x^2-4x+9\)

   • \(f'(x)=14 x - 4\)
   • \(f''(x)=14\)
   • \(f''(2)=14\)

2. (b) \(f(x)=4x^2+6x-3\)

   • \(f'(x)=8 x + 6\)
   • \(f''(x)=8\)
   • \(f''(2)=8\)

Find \(f''(2)\) for each function.

1. (a) \(f(x)=4x^3-x^2+6x+2\)

   • \(f'(x)=12 x^{2} - 2 x + 6\)
   • \(f''(x)=24 x - 2\)
   • \(f''(2)=46\)

2. (b) \(f(x)=x^3+5x^2-8x+11\)

   • \(f'(x)=3 x^{2} + 10 x - 8\)
   • \(f''(x)=6 x + 10\)
   • \(f''(2)=22\)

Find \(y''(2)\) for each function. Give your answers to 3 significant figures.

1. (a) \(y=e^{-3x}\)

   • \(y'(x)=- 3 e^{- 3 x}\)
   • \(y''(x)=9 e^{- 3 x}\)
   • \(y''(2)=\frac{9}{e^{6}}\) = 0.0223

2. (b) \(y=5e^{2x}\)

   • \(y'(x)=10 e^{2 x}\)
   • \(y''(x)=20 e^{2 x}\)
   • \(y''(2)=20 e^{4}\) = 1090

Find \(y''(2)\) for each function. Give your answers to 3 significant figures.

1. (a) \(y=6\ln x\)

   • \(y'(x)=\frac{6}{x}\)
   • \(y''(x)=- \frac{6}{x^{2}}\)
   • \(y''(2)=- \frac{3}{2}\) = −1.5

2. (b) \(y=1-3\ln x\)

   • \(y'(x)=- \frac{3}{x}\)
   • \(y''(x)=\frac{3}{x^{2}}\)
   • \(y''(2)=\frac{3}{4}\) = 0.75

Find the values of x for which \(f(x)=x^3-6x^2+4x+1\) is concave-down.

• \(f'(x)=3 x^{2} - 12 x + 4\)
• \(f''(x)=6 x - 12\)
• \(f''(x)<0\) gives \(6 x - 12<0\)
• \(x<2\)

The graph shows part of the curve \(y=x^3-6x^2+9x\). Use the words increasing, decreasing, concave-up and concave-down to describe each section.

1. (a) A to B

   • A to B: the gradient and the concavity do not change: increasing, concave-down.

2. (b) B to D

   • The curve changes at a labelled point between B and D, so describe each piece.
   • Decreasing; concave-down from B to C, then concave-up from C to D.

3. (c) D to E

   • D to E: the gradient and the concavity do not change: increasing, concave-up.

Water runs into a tank. The volume of water, V litres, t minutes after a tap is opened is modelled by \(V=-0.5t^3+6t^2+20t\).

1. (a) Find \(V''(2)\).

   • \(V'=- \frac{3 t^{2}}{2} + 12 t + 20\)
   • \(V''=12 - 3 t\)
   • \(V''(2)=6\) litres per minute².

2. (b) Say what the sign of your answer tells you about the rate at which the volume is increasing at t = 2.

   • Positive: the gradient is increasing, so the rate at which the volume is increasing is getting larger.

A car bought new is worth \(V=20\,000e^{-0.2t}\) dollars after t years.

1. (a) Find \(V''(2)\) to 2 decimal places.

   • \(V'=-4000e^{-0.2t}\)
   • \(V''=800e^{-0.2t}\)
   • \(V''(2)=536.26\) dollars per year².

2. (b) Is the curve concave-up or concave-down? Say what this means for the model.

   • \(V''>0\) for every t, because \(e^{-0.2t}\) is always positive.
   • The curve is concave-up and decreasing: the car loses value more slowly as it gets older.

Find the values of x for which \(f(x)=x^4-8x^2+3\) is concave-down.

• \(f'(x)=4 x^{3} - 16 x\)
• \(f''(x)=12 x^{2} - 16\)
• \(f''(x)<0\): \(12 x^{2} - 16<0\)
• \(x^2<\frac{4}{3}\), so \(-\frac{2 \sqrt{3}}{3}<x<\frac{2 \sqrt{3}}{3}\)

Find \(d^2y/dx^2\) for \(y=4e^{-x}+3x\), and explain why the curve is concave-up for every value of x.

• \(y'=3 - 4 e^{- x}\)
• \(y''=4 e^{- x}\)
• \(e^{kx}\) is positive for every x, and the number in front of it is positive, so \(y''>0\) for every x.