Question

hw 15 - increasing and decreasing functions section 3.1: problem (1 point) find the critical value(s) of the function $f(x)=17x^{2}e^{-x}$. $x_0 =$ you have attempted this problem 0 times. you have 6 attempts remaining. email instructor

Explanation:

Step1: Apply product - rule for differentiation

The product - rule states that if \(y = uv\), where \(u\) and \(v\) are functions of \(x\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = 17x^{2}\) and \(v = e^{-x}\). Then \(u^\prime=34x\) and \(v^\prime=-e^{-x}\). So \(f^\prime(x)=34x e^{-x}-17x^{2}e^{-x}=17xe^{-x}(2 - x)\).

Step2: Set the derivative equal to zero

To find the critical values, we set \(f^\prime(x) = 0\). Since \(17xe^{-x}(2 - x)=0\), and \(e^{-x}
eq0\) for all real \(x\), we solve \(17x(2 - x)=0\).

Step3: Solve the resulting equation

Setting \(17x(2 - x)=0\), we have two cases: \(17x = 0\) gives \(x = 0\), and \(2 - x=0\) gives \(x = 2\).

Answer:

\(x_0 = 0,2\)