Question

find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)

$f(x)=x^{4}e^{-5x}$

$x=$

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Explanation:

Step1: Differentiate the function using the product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = x^{4}\) and \(v=e^{-5x}\).
\(u^\prime = 4x^{3}\) (using the power rule \((x^{n})^\prime=nx^{n - 1}\)), \(v^\prime=- 5e^{-5x}\) (using the chain rule \((e^{ax})^\prime=ae^{ax}\))
\(f^\prime(x)=4x^{3}e^{-5x}+x^{4}(-5)e^{-5x}=x^{3}e^{-5x}(4 - 5x)\)

Step2: Set the derivative equal to zero

Since \(e^{-5x}>0\) for all \(x\in R\), we set \(x^{3}(4 - 5x)=0\)
\(x^{3}=0\) gives \(x = 0\)
\(4-5x=0\) gives \(x=\frac{4}{5}\)

Answer:

\(0,\frac{4}{5}\)