Graphical Understanding of Derivatives: #5

Question

Determine the intervals on which the given function is concave up or down and find the points of inflection. Let

Inflection Point(s) =

The left-most interval is  , and on this interval  is   ?    (Concave Down,   Concave Up )

The middle interval is  , and on this interval  is   ?    (Concave Down,   Concave Up )

The right-most interval is  , and on this interval  is   ?  (Concave Down,   Concave Up )

Answer 1

For f(x) = (x² - 4)e^x to find the first derivative we use the product rule (uv)' = u'v + uv'

u = x² - 4

u' = 2x

v = e^x

v' = e^x

f '(x) = 2xe^x + (x² - 4)e^x

f '(x) = (x² + 2x - 4)e^x

Although this problem doesn't ask for them, the critical points are found by setting f '(x) = 0 which is only true when x² + 2x - 4 = o so the critical points are x = -1-√5 and x = -1+√5. Using the 1st derivative rule we see that x = -1-√5 is a local maximum and x = -1+√5 is a local minimum.

To find the second derivative, use the product rule again:

u = x² + 2x - 4

u' = 2x + 2

v = e^x

v' = e^x

f ''(x) = (2x + 2)e^x + (x² + 2x - 4)e^x

f ''(x) = (x² + 4x - 2)e^x

To find the points of inflection, we set this equal to zero:

(x² + 4x - 2)e^x = 0 which is true only when:

x² + 4x - 2 = 0

x = -2+√6 and x=-2-√6

These are the points of inflection.

Putting these points of inflection on a number line (shown in red) with the critical points shown (in blue) as a reference, we can check the signs of the second derivative in the three intervals that the inflection points create.

On the interval (-∞, -2-√6) the second derivative is positive meaning the function is concave up.

On the interval (-2-√6, 2-√6) the second derivative is negative meaning the function is concave down.

On the interval (2-√6, ∞) the second derivative is positive meaning the function is concave up.