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Critical points?

2 Answers

(1)

Step 1:

The function is $\\f\left ( x \right )=\left (e^{x}+e^{-x} \right ).$

Let $\\f\left ( x \right )=y$ .

$y=e^{x}+e^{-x}\rightarrow\left ( 1 \right )$

$\\f\left ( x \right )=\left (e^{x}+e^{-x} \right )$

Differentiate on each side with respect to x.

$\\f'\left ( x \right )=\frac{d}{dx}\left (e^{x}+e^{-x} \right )$

Sum rule of derivatives : .

$f'\left ( x \right )=\frac{d}{dx}\left \left (e^{x} \right )+\frac{d}{dx}\left \left (e^{-x} \right )$

$f'\left ( x \right )=\left (e^{x} \right ) \right )+\left \left (e^{-x} \right )\frac{d}{dx}\left ( -x \right )$

$f'\left ( x \right )=\left e^{x} -e^{-x}$ .

Step 2:

To find the critical numbers of $f\left ( x \right )$ , equate $f'\left ( x \right )$ to zero.

$f'\left ( x \right )=0$ .

$e^{x} -e^{-x}=0$

$e^{x} -\frac{1}{e^{x}}=0$

$\\e^{2x}-{1} =0\\e^{2x}=1.$

Apply logarithm on each side.

$\\\ln e^{2x}=\ln 1 \\2x=0\\x=0.$

Substitute $x=0$ in equation (1).

$\\y=e^{0}+e^{\left (-0 \right )}$

$y=2$ .

Critical values are : $x=0$ and $y=2$ .

Solution :

Critical values are : $x=0$ and $y=2$ .

answered Mar 18, 2015 by sandy Pupil

(2)

Step 1:

The function is $\\g\left ( x \right )=x^{3}\ln x$ .

The domain of the function is $(0, \infty )$ .

Let $\\g\left ( x \right )=y$ .

$\\y=x^{3}\ln x \rightarrow \left ( 1 \right )$

$\\g\left ( x \right )=x^{3}\ln x$

Differentiate with respect to x on each side.

$\\g'\left ( x \right )=\frac{d}{dx}\left (x^{3}\ln x \right )$

Product rule of derivatives : $\frac{d}{dx}\left \left (uv \right )=u\frac{dv}{dx}+v\frac{du}{dx}.$

$g'\left ( x \right )=x^{3}\frac{d}{dx}\left ( \ln x \right )+\left (\ln x \right )\frac{d}{dx}\left ( x^{3} \right )$

$g'\left ( x \right )=x^{3}\left (\frac{1}{x} \right )+\left (\ln x \right )\left ( 3x^{2} \right )$

$\\g'\left ( x \right )=x^{2}+ 3x^{2} \right \ln x\\g'\left ( x \right )=x^{2}\left ( 1+3\ln x \right )$ .

Step 2:

To find the critical numbers of $g\left ( x \right )$ , equate $g'\left ( x \right )$ to zero.

$x^{2}\left ( 1+3\ln x \right )=0$

$x^{2}=0$ and $(1+3\ln x) =0$

$x=0$ and $(1+3\ln x) =0$ .

since $x=0$ is not in the domain, $x=0$ it is not considered.

$\ln x =\frac{-1}{3}$

$x =e^{\frac{-1}{3}}$

$x=0.71$ .

Substitute $x=0.71$ in equation (1).

$\\y=\left (0.71 \right )^{3}\ln \left (0.71 \right )$

$y=-0.122.$

Critical values are : $x=0.71$ and $y=-0.122.$

Solution :

Critical values are : $x=0.71$ and $y=-0.122.$

answered Mar 18, 2015 by sandy Pupil
edited Mar 18, 2015 by sandy

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