horizontal tangent line for f(x) = (x^3)/(e^x) ?

tinad

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Jun 4, 2006
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If e^x is 0 and the original function is f(x) = x^3/e^x, can there be a horizontal tangent line?

i know that the degree of the numerator is higher than that of the denominator so the graph of f has no horizontal
 
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.
 
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.
 
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