Question

Find the critical points, the second derivative, and local minimum and maximum

Answer 1

Recall that a critical point is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero.

1) The first derivative of the given function is:

`\begin{gathered} f^(\prime)(x)=-e^x(x-9)+(-e^x)*1 \\ =-e^x(x-9+1)=-e^x(x-8). \end{gathered}`

Notice that f'(x) is the product of two continuous functions, then it is continuous.

Setting f'(x)=0 we get:

`0=-e^x(x-8).`

Then:

`e^x=0\text{ or }x-8=0.`

We know that for all x real number:

`e^x>0.`

Then:

`\begin{gathered} x-8=0, \\ x=8. \end{gathered}`

Answer 1:

Option A) The critical point(s) is(are) x=8.

2) Now, notice that f''(x) is the derivative of f'(x), then:

`\begin{gathered} f^(\prime)^(\prime)(x)=(-e^x(x-8))^(\prime)=-e^x(x-8)+(-e^x*1) \\ =-e^x(x-7). \end{gathered}`

Now, evaluating f''(x) at x=8 we get:

`f^(\prime)^(\prime)(8)=-e^8(8-7)=-e^8<0.`

Using the second derivative test we get that f(x) reaches a local maximum.

Answer 2:

`f^(\prime)^(\prime)(x)=-e^x(x-7).`

f(x) reaches a local maximum at x=8.