Given a function, find the critical point.

Question

Let f(x) = 6bxe^1+bt , where b > 0.

I found the derivative, and that was right in my online grading system. I then proceeded to set the derivative equal to zero and solve for the critical point. I got 1/b but that was marked wrong. Having variables and not numbers is realy confusing me and I was wondering if you could help me figure out where I went wrong.

Answer 1

Lets set the derivative equal to zero.  Use product rule and chain rule to derive.

 

6be^{1 + bt} + 6bx(b)e^{1 + bt}= 0

 

6be^{1 + bt} + 6b²xe^{1 + bt} = 0

 

6be^{1 + bt}(1 + bx) = 0

 

1 + bx = 0

       bx = -1

         x = -1/b

 

You missed a negative sign in the critical value.

Answer 2

f'(x) = 6be^1+bt   can never be zero.