# Calculus BC: Applications of the Derivative

### Contents

#### Problems

Problem : Find the critical points and inflection points of the function f (x) = x4 -2x2 (with domain the set of all real numbers). Which of the critical points are local minima? local maxima? Is there a global minimum or maximum?

We first calculate the derivatives of the function:

 f'(x) = 4x3 - 4x = 4(x + 1)x(x - 1) f''(x) = 12x2 - 4 = 4(3x2 - 1)

We see that f'(x) = 0 when x = - 1, 0, or 1, so these are the three critical points of f. We calculate the second derivatives at these points:

 f''(- 1) = 8 f''(0) = -4 f''(1) = 8

so by the second derivative test, f has local minima at -1 and 1 and a local maximum at 0. Substituting back into the original function yields

 f (- 1) = -1 f (0) = 0 f (1) = -1

so f attains its global minimum of -1 at x = ±1. It is clear from the graph of f that it has no global maximum.
Figure %: Graph of f (x) = x4 -2x2
To find the points of inflection, we solve f''(x) = 0, or 12x2 - 4 = 0, which has solutions x = ±1/3) ±0.58. Referring once again to the graph of f, we can check that the concavity does indeed change at these x-values.

Problem : Find the inflection points of f (x) = e-x2. (This famous function is called the gaussian.)

We compute the derivatives:

 f'(x) = -2xe-x2, f''(x) = (- 2x)(- 2xe-x2) + (- 2)(e-x2) = (4x2 -2)e-x2

Solving f''(x) = 0 for x, we get the inflection points x = ±1/ 0.71. This is reasonable when one considers the graph of f.
Figure %: Graph of f (x) = e-x2

Problem : Find a function f (x) with inflection points at x = 1 and x = 2.

We need f''(1) = f''(2) = 0, so we might as well let

 f''(x) = (x - 1)(x - 2) = x2 - 3x + 2

We search for a function with this derivative, ultimately obtaining

 f'(x) = - + 2x

Similarly, a function with this derivative is given by

 f (x) = - + x2