Second Derivatives and Beyond Exercises

Answer

This function doesn't have a global max or a global min. It keeps increasing in one direction and decreasing in the other:

Answer

This function does have both a global max and a global min:

Answer

This function has a local min and a local max, but doesn't have a global min or a global max.

Answer

This function has a global min, but no global max

Answer

This function only has one value. Since y = 2 is both the biggest and the smallest this function ever becomes, y = 2 is both the global maximum and the global minimum.

Answer

This function has no global min or global max.

Answer

This function is trying to have a global max, but never quite reaches it. It also has no global minimum, since the function approaches zero as x approaches ± ∞, but never reaches zero.

Answer

This function has no global max or global min.

Answer

Rewrite the function to make life easier:

The derivative is

f'(x) = -(x² + 1) ^{- 2}(2x)

which is zero when x = 0.

Now evaluate f at the critical point and the endpoints of the interval:

The global max is y = 1 and the global min is y = 0.5.

Answer

The derivative is

f '(x) = 2x + 3

which is zero when x = -3/2.

We evaluate f at this critical point and at the endpoints:

On the specified interval, the global max of f is y = 18 and the global min of f is y = -2.25.

Answer

The derivative is f '(x) = xe^x + e^x = e^x(x + 1). The only critical point occurs at x = -1, which is also an endpoint of the interval in question. It turns out that we only need to evaluate f (x) at the endpoints.

The global maximum of f on this interval is -2e^-2 and the global minimum is -e^-1.

Answer

The derivative of f is f '(x) = cos x, whose only root on the given interval occurs at .

We evaluate f at the critical point and at the endpoints of the interval:

The global max of f on this interval is y = 0 (which occurs at two different values of x) and the global min is y = -1.

Answer

The derivative is f '(x) = x² + 3x – 10, which factors as

f '(x) = (x + 5)(x – 2).

The critical points occur at x = -5 and at x = 2.

We evaluate f at the critical points and at the endpoints of the interval:

The global max is y = 46.5 and the global min is .

Answer

The derivative is

f '(x) = 2xe^x + x²e^x = xe^x(2 + x)

so there are critical points at x = -2 and at x = 0.

We evaluate f at the critical points and endpoints (it's ok to use approximate y-values here, because we're only comparing the y-values to figure out which x-values we want):

f has its global min when x = 0 and its global max when x = 1.

Answer

The derivative is

f '(x) = 4x + 3

which is zero when , so this is our critical point.

We evaluate f at all the important places:

The global max of f occurs when x = 2 and the global min when .

Answer

The derivative is

which has a critical point whenever

This occurs when

where k is an integer. Solving,

and so

We want only the critical points in the interval [-π,π]. These would be

Now we evaluate f at lots of places:

The global max occurs at and at . The global min occurs at and at .

Answer

The derivative is

This is 0 when -x³ = -1, or when x = 1. This is our only critical point, since anywhere f ' is undefined, f is already undefined.

Now evaluate f at the critical points and endpoints.

The function f achieves its global max on this interval when x = 1 and its global min when x = -1.

Answer

This function is a line, so it has no critical points.

Better yet, we don't even need to evaluate f at the endpoints. Since f is a line with a slope of 3, f is an increasing function. This means f must be smallest at the start of the interval and largest at the end of the interval.

The global min of f on this interval occurs at x = -5 and the global max of f on this interval occurs at x = 5.

These problems become more complicated when we start doing something called optimization, which is the same exact math but with more words added on top.