SOLUTION: This is from my derivatives of exponents lesson. If f(x)=10^(3x-5) times e^[(2x)^2], determine the values of x so that f'(x)=0. Thank you!

Algebra ->  Equations -> SOLUTION: This is from my derivatives of exponents lesson. If f(x)=10^(3x-5) times e^[(2x)^2], determine the values of x so that f'(x)=0. Thank you!      Log On


   

Question 1200978: This is from my derivatives of exponents lesson.
If f(x)=10^(3x-5) times e^[(2x)^2], determine the values of x so that f'(x)=0.
Thank you!
Found 2 solutions by psbhowmick, ikleyn:
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+10%5E%283x-5%29e%5E%282x%29%5E2

d%28f%28x%29%29%2Fdx+=+10%5E%283x-5%29e%5E%282x%29%5E2%284x%2B3ln%2810%29%29
d%28f%28x%29%29%2Fdx+=+10%5E%283x-5%29e%5E%282x%29%5E2%284x%2Bln%281000%29%29

Clearly, d%28f%28x%29%29%2Fdx will be zero only if 4x%2Bln%281000%29 is zero.
4x%2Bln%281000%29+=+0
x+=+%28-1%2F4%29ln%281000%29+=+-1.7269

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

        This my post is a correction to the solution by @psbhowmick.


f(x) = 10%5E%283x-5%29e%5E%28%282x%29%5E2%29 = 10%5E%283x-5%29e%5E%284x%5E2%29.

d%28f%28x%29%29%2Fdx = 10%5E%283x-5%29e%5E%284x%5E2%29%2A8x + 10%5E%283x-5%293ln%2810%29%2Ae%5E%284x%5E2%29

d%28f%28x%29%29%2Fdx = 10%5E%283x-5%29e%5E%28%282x%29%5E2%29%2A%288x%2B3ln%2810%29%29

d%28f%28x%29%29%2Fdx = 10%5E%283x-5%29%2Ae%5E%28%282x%29%5E2%29%2A%288x%2Bln%281000%29%29


        Clearly,   d%28f%28x%29%29%2Fdx   will be zero only if   8x%2Bln%281000%29   is zero.


8x + ln(1000) = 0

x = %28-1%2F8%29ln%281000%29 = -0.86346941.             ANSWER

Solved.