Second Derivatives and Beyond Exercises

Answer

We use the quotient rule to find the derivative:

This is defined for all x in the domain of f, that is, all x greater than zero. It's zero when

ln x = 1

which means when

x = e.

Since the denominator of the derivative is (ln x)², which is always positive, the sign of f '(x) is determined by the sign of the numerator

ln x - 1.

This is negative when x < e and positive when x > e, so we find this numberline:

Therefore f (x) is decreasing to x = e and then increasing, so f (x) hits a minimum at x = e.

Answer

The derivative is

This is only zero when x = 1, and never undefined, so x = 1 is the only critical point. Since the quantity (x – 1)² is positive for all x ≠ 1, the derivative

f '(x) = -3(x – 1)²

is negative for all x ≠ 1. We find this numberline:

Since f is decreasing, levels out, and then decreases again, we have neither a minimum nor a maximum at x = 1.

Answer

We use the quotient rule to find the derivative:

f '(x) is undefined only at x = -3, in which case f is also undefined so this is not a critical point. f '(x) is zero when

x² + 6x + 5 = 0.

Happily, this quadratic factors as

x² + 6x + 5 = (x + 5)(x + 1),

so f '(x) is zero at x = -5 and x = -1. Here's the numberline so far:

To figure out whether f has a maximum or a minimum at each of these critical points, we need to find the sign of the first derivative so we can fill in the numberline.

The derivative

is negative when exactly one of the quantities (x + 5) or (x + 1) is negative. This is only possible when

-5 < x < -1,in which case (x + 1) is negative but (x + 5) is positive. So the numberline looks like this:

Since f ' is positive to the left and negative to the right of x = -5, the function f has a maximum at x = -5. Since f ' is negative to the left and positive to the right of x = -1, the function f has a minimum at x = -1.

Answer

We use the quotient rule to find the derivative:

This is only undefined when x² – 3x + 2 = 0, but such an x value doesn't count as a critical point since it also makes f undefined. The derivative is zero when the numerator is zero, which occurs when

Now we have this numberline:

Since the denominator of f ' is always positive, the sign of f ' is the same as the sign of its numerator. The numerator

-6x² + 12

is positive when 6x² is smaller than 12, that is, when x² is smaller than 2.

On the numberline, we have this:

Since f ' is negative to the left of x =  and positive to the right of x = , f has a minimum at x = . Since f ' is positive to left of x =  and negative to the right, f has a maximum at x = .

Answer

The derivative of f (x) is

f '(x) = e^x.

Since f '(x) is always positive, the function f (x) has no critical points and therefore cannot have any maxima or minima.