Is $f(x)=x^2e^x$ increasing or decreasing at $x=1$ ?

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Answer 1 · 2016-08-30T01:15:47

To know that, you are finding the slope at $x = 1$ , and seeing if it is positive or negative (or $0$ , but in that case it would be an extremum).

$(df)/(dx) = d/(dx)[x^2e^x]$

$= x^2e^x + 2xe^x$ (product rule)

$= e^x (x^2 + 2)$

Thus, at $x = 1$ :

$f(1) = e^(1) ((1)^2 + 2)$

$= 3e > 0$

So, the function is increasing at $x = 1$ .

Is f(x)=x^2e^x f(x)=x^2e^x increasing or decreasing at x=1 x=1 ?