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

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

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Answer 1 · 2016-08-30T01:15:47

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

$\frac{d f}{d x} = \frac{d}{d x} [x^{2} e^{x}]$ $(df)/(dx) = d/(dx)[x^2e^x]$

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

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

Thus, at $x = 1$ $x = 1$ :

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

$= 3 e > 0$ $= 3e > 0$

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

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

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Is f(x)=x2exf(x)=x^2e^x increasing or decreasing at x=1x=1 ?

1 Answer

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Impact of this question

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