Second Derivatives and Beyond Exercises

Answer

First we find the derivative, using the quotient rule:

We're leaving the derivative in this form, with the numerator and denominator factored, because it will make the next pieces of the task easier.

Next we need to find the x-values at which the derivative is undefined but f is defined. Since f '(x) is a rational function, it's undefined wherever the denominator is 0.

To have

(x + 2)² = 0

we need to have

x + 2 = 0

or

x = -2,

so f ' is undefined at x = -2. However, f is also undefined at x = -2, so x = -2 is not a critical point. Finally, we need to find the roots of f '. Since f '(x) is a rational function, it will be 0 when the numerator is 0 and the denominator isn't 0.In order to have

x(x + 4) = 0

we must have either x = 0 or x = -4. Since neither of these values make the denominator of f ' equal zero, x = 0 and x = -4 are critical points. If we look at the graph of f, this makes sense:

Answer

f (x) = cos x

The derivative is

f '(x) = -sin x.

This is never undefined, and is 0 at all multiples of π. So the critical points of f (x) are x = nπ, where n is any integer. If we look at a graph of f, this matches our intuition:

Answer

We find the derivative using the quotient rule.

The derivative f '(x) is undefined when x ≤ 0, but since we're only looking at the function f for x > 0, we don't find any critical points because f '(x) is undefined. To find where f '(x) = 0, we find where the numerator is equal to 0.

If

ln x – 1 = 0

then

ln x = 1,

therefore

x = e.

This is our only critical point.

Answer

Again, we need to use the quotient rule to find the derivative. Be careful when factoring out the negative signs.

Now we can investigate where f '(x) is undefined or zero. f '(x) is undefined when x = 0. However, this doesn't count as a critical point since x = 0 wasn't in the domain of f to begin with. f '(x) is zero when

-28(x + 1) = 0,

which is when

x = -1.

This is our only critical point.

If we look at a graph, we can see this critical point, but we need to zoom in a lot:

We need to be careful with the calculator. Make sure it's telling the correct answer.

Answer

f (x) = xe^x

We use the product rule to find the derivative:

This function is never undefined. Since e^x is never zero, x = -1 is the only root of f '(x) and therefore the only critical point.

Again, this makes sense on the graph:

Answer

f (x) = 4x³ + 3x² – 2x + 1

We find the derivative:

f '(x) = 12x² + 6x – 2.

Since this is a polynomial, it's defined everywhere. In order to find the roots of the derivative, first factor out the unnecessary 2:

f '(x) = 2(6x² + 3x – 1).

Then we need to use the quadratic formula:

These are the critical points, and our exact answers. However, to check our work on the graph it's useful to know approximate values:

If we graph the function f using marks every 0.25 on the x-axis, we do appear to have horizontal tangents around 0.25 and -0.75. This is enough evidence to support our answer.

Answer

f (x) = sin x cos x

We find the derivative using the product rule.

This is never undefined. f '(x) is zero when

cos² x = sin² x,

or when

cos x = ± sin x.

Think about the unit circle. sin and cos are equal at π/4, 3π/4, 5π/4, 7π/4, and anything you find by adding or subtracting 2π to any of these numbers. Look at a graph and see if this is believable:

Yup, it's believable. We have horizontal tangent lines everywhere we expected to. So our critical points occur at

where k is any integer.

Answer

f (x) = sin² x

We can rewrite the function as

f (x) = (sin x)²,

which means we need the chain rule to take the derivative of this function.

f '(x) = 2(sin x)(cos x)

The derivative f '(x) is never undefined, but is 0 whenever either sine or cosine is 0—that is, at every multiple of π/2. So these are our critical points. The graph looks like this:

Answer

f (x) = e^x + e^-x

We take the derivative (using the chain rule for the second term).

f '(x) = e^x – e^-x.

This is never undefined, and is zero when

e^x = e^-x.

But this is only possible when x = -x, which means

x = 0

is our only critical point.

Answer

f (x) = 3x – 2^x

We find the derivative:

f'(x) = 3 – 2^x ln 2.

This is never undefined. Find its root(s):

That horrible expression is the exact value of the critical point. For the sake of checking the graph, this value is approximately 2.1.