Second Derivatives and Beyond Exercises

Answer

f (x) = e^x

This function is never 0, so there are no x-intercepts. The y-intercept is (0, 1). For the sake of future graphing, notice this means f (x) is always positive since it can't cross the x-axis to become negative.

The derivative is f '(x) = e^x. Since this is never zero or undefined, there are no critical points.

The second derivative is f "(x) = e^x. Since this is never zero or undefined, there are no possible inflection points.

Answer

f (x) = (x – 2)³

This function is 0 when x = 2, so the x-intercept is (2, 0). The y-intercept is (0, -8).

The derivative is

f '(x) = 3(x – 2)²

which is also zero when x = 2, so the x-intercept (2, 0) is also a critical point.

The second derivative is

f"(x) = 6(x – 2) which is also zero when x = 2. Since f " changes sign at x = 2, (2, 0) is also an inflection point.

Answer

f (x) = (1 – x)e^x

The y-intercept is (0, 1). The x-intercept is (1, 0).

The derivative is

f '(x) = (-1)e^x + (1 – x)e^x = -xe^x.

This is zero when x = 0, so (0,1) is a critical point as well as an intercept.

The second derivative is

f "(x) = -e^x – xe^x = -e^x(1 + x).

This is zero when x = -1 and changes sign at x = -1, so we have an inflection point at

For the sake of graphing, it's useful to know that

Answer

f (x) = x²e^x – 4e^x

Rewrite f so we can tell what's going on.

f (x) = e^x(x² – 4) = e^x(x – 2)(x + 2).

This is zero when x = ± 2, so the x-intercepts are (2, 0) and (-2, 0). The y-intercept is (0,-4).

The derivative of f is

We need to use the quadratic formula to find the roots of (x² + 2x – 4), which will be the only places where f ' is zero.

The critical points are

and

The second derivative of f is

Again, we need to use the quadratic formula. We need to find the roots of (x² + 4x – 2), since these are the places f " will be zero.

Since the sign of f" does change at these x-values, these are both inflection points of f. We will find their full coordinates.

We need to cheat a little on the labeling, because the exact coordinates are so awful.

Answer

The function f has an x- and y-intercept of (0,0), and a vertical asymptote at x = -2.

The derivative is

This is undefined at x = -2, which isn't in the domain of f anyway, so f has no critical points.

The second derivative is

(you can find this using the quotient rule or by re-writing f ' and using the chain rule). The second derivative is also undefined at x = -2, and not zero anywhere, so f has no inflection points.