Consider the following function.

Question

f(x) = (3 − x)e^−x

a) Find the intervals of increase or decrease. (Enter your answers using interval notation.)

increasing
decreasing

(b) Find the intervals of concavity. (Enter your answers using interval notation. If an answer does not exist, enter DNE.)

concave up
concave down

(c) Find the point of inflection. (If an answer does not exist, enter DNE.)

(x, y) =  

Answer 1

f(x) = (3 - x)e^-x

f'(x) = -e^-x - (3-x)e^-x = -e^-x(1 + 3 - x) = -e^-x(4 - x) = 0 when x = 4

When x < 4, f'(x) < 0, so f is decreasing

When x . 4, f'(x) > 0, so f is increasing

f"(x) = e^-x(4 - x) + e^-x = e^-x(4 - x + 1) = e^-x(5 - x) = 0 when x = 5

When x < 5, f"(x) > 0. So f is concave up.

When x > 5, f"(x) < 0. So f is concave down

Inflection point when x = 5.

Answer 2

Take the derivative and set it equal to zero to find the places where the function has a tangent line equal to zero (i.e., the local max's or min's). These will be the places where the function MAY change from increasing to decreasing or visa-versa. To take the derivative, you must use the product rule:

(u•v)' = u'v + uv'

u = (3 - x)

u' = -1

v = e^-x

v' = (-1)e^-x = -e^-x

To find the concavity, take the second derivative. For values of "x" that make f'' negative, the function will be concave down and values of "x" that make f'' positive, the function will be concave up. To find those "dividing lines", set the second derivative equal to zero and solve. These will be places where the concavity MAY change. Again, you'll need to use the product rule to take the derivative.