# the function f(x) = e^3x +6x^2 +1 has a horizontal tangent at x =?

• February 25th 2009, 04:08 PM
LexiRae
the function f(x) = e^3x +6x^2 +1 has a horizontal tangent at x =?
i cant solve it!
im really lost. i know you find f''(x) (i got 9e^3x +12) but i dont know where to go from there
what would i do on my calculator?
• February 25th 2009, 04:34 PM
Reckoner
Quote:

Originally Posted by LexiRae
i cant solve it!
im really lost. i know you find f''(x) (i got 9e^3x +12) but i dont know where to go from there
what would i do on my calculator?

A horizontal tangent will have a slope of 0. So set $f'(x)=0$ and solve for $x.$
• February 25th 2009, 04:53 PM
skeeter
Quote:

Originally Posted by Reckoner
A horizontal tangent will have a slope of 0. So set $f''(x)=0$ and solve for $x.$

set $f'(x) = 0$ rather than $f''$ ?

$f(x) = e^{3x} + 6x^2 + 1
$

$f'(x) = 3e^{3x} + 12x = 0$

graph f'(x) and look for the zero.

using a calculator ... $x \approx -0.156$