How do you find local maximum value of f using the first and second derivative tests: $f (x) = x^{3} - 2 x + 5$ $f(x)=x^3-2x+5$ on the interval (-2,2)?

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Answer 1 · 2017-11-27T22:04:44

See below.

First differentiate $x^{3} - 2 x + 5$ $x^3-2x+5$

$\frac{d}{d x} (x^{3} - 2 x + 5) = 3 x^{2} - 2$ $d/dx(x^3-2x+5)=3x^2-2$

equating this to zero and solving fot $x$ $x$ will give us the points where the tangent is horizontal .i.e. maximum and minimum values.

$3 x^{2} - 2 = 0 \Rightarrow x = \pm \sqrt{\frac{2}{3}}$ $3x^2-2=0=>x= +-sqrt(2/3)$

We now use the second derivative test. If:

$\frac{d^{2} y}{{d x}^{2}} > 0$ $(d^2y)/(dx^2)>0$ minimum value.

$\frac{d^{2} y}{{d x}^{2}} < 0$ $(d^2y)/(dx^2)<0$ maximum value.

$\frac{d^{2} y}{{d x}^{2}} = 0$ $(d^2y)/(dx^2)=0$ min/max or point of inflection.

Second derivative:

$\frac{d^{2} y}{{d x}^{2}} (3 x^{2} - 2) = 6 x$ $(d^2y)/(dx^2)(3x^2-2)=6x$

$6 (+ \sqrt{\frac{2}{3}}) > 0$ $6(+sqrt(2/3))>0$ min value.

$6 (- \sqrt{\frac{2}{3}}) < 0$ $6(-sqrt(2/3))<0$ max value.

These are both in the interval $(- 2, 2)$ $( -2 , 2 )$

$x = - \sqrt{\frac{2}{3}}$ $x=-sqrt(2/3)$ give maximum value so:

$f (- \sqrt{\frac{2}{3}}) = {(- \sqrt{\frac{2}{3}})}^{3} - 2 (- \sqrt{\frac{2}{3}}) + 5$ $f(-sqrt(2/3))=(-sqrt(2/3))^3-2(-sqrt(2/3))+5$

$\to = \frac{4 \sqrt{6}}{9} + 5 \approx 6.089$ $->=(4sqrt(6))/9+5~~6.089$

Graph of function and first derivative:

enter image source here

How do you find local maximum value of f using the first and second derivative tests: f(x)=x^3-2x+5f(x)=x3−2x+5f(x)=x^3-2x+5 on the interval (-2,2)?