Calculus + Critical Number

Question

The function f(x)= (6x-3)e^5x has one critical number.

 

What is the critical number?

Answer 1

f'(x) = 6e^5x + (6x-3)e^5x·5

 

= [6 + 5(6x - 3)]e^5x

 

= (30x - 9)e^5x

 

= 3(10x - 3)e^5x

 

This is 0 when 10x - 3 = 0, so there is only one critical point, which is at x = 3/10,

 

Here, the function has the value [6(3/10) - 3]e^5(3/10) = -6e^3/2/5 ≈ 5.378

 

f''(x) =30e^5x + (30x - 9)e^5x·5

 

= [30 + 5(30x - 9)]e^5x

 

= (150x - 15)e^5x

 

= 15(10x - 1)e^5x

 

f''(3/10) = 15[10(3/10) - 1]e^5(3/10) = 30e^5/2 > 0.

 

So this is a local minimum. In fact, it is a global minimum.