Calc I question HELP!!!

Question

Let f(x) = 9xe^x

(a) Find the exact value of x so that
f ''(x)=0.
so x= -2

I am having trouble solving the next question:
(b) For what interval is
f(x) concave up? (Enter your answer using interval notation.)

Answer 1

Two ways to look at it:

 

1. The second derivative here is continuous, and has just one zero. Therefore there are two intervals to consider, (-infty,-2) and (-2,infty). Try plugging in any one point from each interval to learn if it is positive or negative on the entire interval. Since it doesn't cross zero anywhere else, it can't switch signs within that interval.

 

2. Note that

x < -2 implies x+2 < 0, which

implies 9e^x(x+2) < 0. So

9e^x( x+2 ) < 0 on (-infty,-2) and likewise > 0 on (-2,infty) simply by algebra.

 

 

With either method, the theorem states that if f'' is positive on an interval then f is concave up on the interval, and similarly negative means concave down.

Answer 2

a)

 

Find the second derivative of f(x) first.  Then set that second derivative equal to zero.

 

f'(x) = 9e^x + 9xe^x

 

f''(x) = 9e^x + 9e^x + 9xe^x

 

f''(x) = 18e^x + 9xe^x

 

f''(x) = 9e^x(2 + x)

 

 

9e^x(2 + x) = 0

 

2 + x = 0

 

x = -2

 

 

You have this correct.

 

 

b)

 

Evaluate the second derivative at x=-3 and x=-1. 

 

If the second derivative is positive at x=-3, then concave up in the interval (-∞, -2).

If the second derivative is positive at x=-1, then concave up in the interval (-2, ∞).

If the second derivative is positive in both cases, then concave up in the interval (-∞, ∞).