The second derivative and concavity Problem set B Worked solutions

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

1. (a) \(f(x)=9x^2+5x-2\)

   • \(f'(x)=18 x + 5\)
   • \(f''(x)=18\)
   • \(f''(2)=18\)

2. (b) \(f(x)=3x^2-8x+7\)

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

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

1. (a) \(f(x)=3x^3+2x^2-5x+8\)

   • \(f'(x)=9 x^{2} + 4 x - 5\)
   • \(f''(x)=18 x + 4\)
   • \(f''(2)=40\)

2. (b) \(f(x)=x^3-4x^2+9x+6\)

   • \(f'(x)=3 x^{2} - 8 x + 9\)
   • \(f''(x)=6 x - 8\)
   • \(f''(2)=4\)

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

1. (a) \(y=e^{-4x}\)

   • \(y'(x)=- 4 e^{- 4 x}\)
   • \(y''(x)=16 e^{- 4 x}\)
   • \(y''(2)=\frac{16}{e^{8}}\) = 0.00537

2. (b) \(y=3e^{2x}\)

   • \(y'(x)=6 e^{2 x}\)
   • \(y''(x)=12 e^{2 x}\)
   • \(y''(2)=12 e^{4}\) = 655

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

1. (a) \(y=4\ln x\)

   • \(y'(x)=\frac{4}{x}\)
   • \(y''(x)=- \frac{4}{x^{2}}\)
   • \(y''(2)=-1\)

2. (b) \(y=2-5\ln x\)

   • \(y'(x)=- \frac{5}{x}\)
   • \(y''(x)=\frac{5}{x^{2}}\)
   • \(y''(2)=\frac{5}{4}\) = 1.25

Find the values of x for which \(f(x)=x^3-9x^2+7x+4\) is concave-down.

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

The graph shows part of the curve \(y=-x^3+6x^2-9x+6\). 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: decreasing, concave-up.

2. (b) B to D

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

3. (c) D to E

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

The height of a bean plant, h cm, t weeks after planting is modelled by \(h=0.5t^3-6t^2+30t\).

1. (a) Find \(h''(3)\).

   • \(h'=\frac{3 t^{2}}{2} - 12 t + 30\)
   • \(h''=3 t - 12\)
   • \(h''(3)=-3\) cm per week².

2. (b) Say what the sign of your answer tells you about the rate at which the plant is growing at t = 3.

   • Negative: the gradient is decreasing, so the rate at which the plant is growing is getting smaller.

The concentration of a drug in the blood is \(C=60e^{-0.3t}\) mg per litre, t hours after a dose.

1. (a) Find \(C''(2)\) to 3 significant figures.

   • \(C'=-18e^{-0.3t}\)
   • \(C''=5.4e^{-0.3t}\)
   • \(C''(2)=2.96\) mg per litre per hour².

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

   • \(C''>0\) for every t, because \(e^{-0.3t}\) is always positive.
   • The curve is concave-up and decreasing: the concentration falls more slowly as time passes.

Find the values of x for which \(f(x)=x^4-12x^2+5\) is concave-down.

• \(f'(x)=4 x^{3} - 24 x\)
• \(f''(x)=12 x^{2} - 24\)
• \(f''(x)<0\): \(12 x^{2} - 24<0\)
• \(x^2<2\), so \(-\sqrt{2}<x<\sqrt{2}\)

Find \(d^2y/dx^2\) for \(y=2e^{3x}-5x\), and explain why the curve is concave-up for every value of x.

• \(y'=6 e^{3 x} - 5\)
• \(y''=18 e^{3 x}\)
• \(e^{kx}\) is positive for every x, and the number in front of it is positive, so \(y''>0\) for every x.