Is $f (x) = e^{x} \sqrt{x^{2} - x}$ $f(x)=e^xsqrt(x^2-x)$ increasing or decreasing at $x = 3$ $x=3$ ?

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Answer 1 · 2016-06-25T06:32:24

$f (x) = e^{x} \sqrt{x^{2} - x}$ $f(x)=e^xsqrt(x^2-x)$ is increasing at $x = 3$ $x=3$

A function $f (x)$ $f(x)$ is increasing at $x = a$ $x=a$ if $f'(a)>0$ and is decreasing at $x=a$ if $f'(a)<0$ .

As $f(x)=e^xsqrt(x^2-x)$ , using product rule

$f'(x)=e^xsqrt(x^2-x)+e^x xx1/(2sqrt(x^2-x))xx(2x-1)$

= $e^xsqrt(x^2-x)+(e^x(2x-1))/(2sqrt(x^2-x))$

and at $x=3$

$f'(x)=e^3sqrt(3^2-3)+(e^3(2xx3-1))/(2sqrt(3^2-3))$

= $e^3sqrt6+(5e^3)/(2sqrt6)=e^3sqrt6(1+5/12)=e^3sqrt6xx17/12$

which is clearly positive.

Hence, $f(x)=e^xsqrt(x^2-x)$ is increasing at $x=3$ .

Is f(x)=e^xsqrt(x^2-x)f(x)=ex√x2−xf(x)=e^xsqrt(x^2-x) increasing or decreasing at x=3x=3x=3?