horizontal tangent line for f(x) = (x^3)/(e^x) ?
- Thread starter tinad
- Start date
ChaoticLlama
Junior Member
- Joined
- Dec 11, 2004
- Messages
- 199
Sorry, misread question.
However, I question your statement " if e^x = 0 "
Solve for x in that statment please, and tell me what you find.
However, I question your statement " if e^x = 0 "
Solve for x in that statment please, and tell me what you find.
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,204
a function has a horizontal tangent (which is not the same as a horizontal asymptote) wherever f'(x) = 0.
\(\displaystyle f(x) = \frac{x^3}{e^x}\)
\(\displaystyle f'(x) = \frac{e^x 3x^2 - x^3 e^x}{e^{2x}}\)
\(\displaystyle f'(x) = \frac{x^2 e^x(3 - x)}{e^{2x}}\)
\(\displaystyle f'(x) = \frac{x^2(3 - x)}{e^x} = 0\)
f'(x) = 0 at x = 0 and x = 3.
\(\displaystyle f(x) = \frac{x^3}{e^x}\)
\(\displaystyle f'(x) = \frac{e^x 3x^2 - x^3 e^x}{e^{2x}}\)
\(\displaystyle f'(x) = \frac{x^2 e^x(3 - x)}{e^{2x}}\)
\(\displaystyle f'(x) = \frac{x^2(3 - x)}{e^x} = 0\)
f'(x) = 0 at x = 0 and x = 3.