How do you find the local maximum and minimum values of $f (x) = 5 + 9 x^{2} - 6 x^{3}$ $f(x) = 5 + 9x^2 − 6x^3$ using both the First and Second Derivative Tests?

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How do you find the local maximum and minimum values of $f (x) = 5 + 9 x^{2} - 6 x^{3}$ $f(x) = 5 + 9x^2 − 6x^3$ using both the First and Second Derivative Tests?

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Answer 1 · 2017-08-04T17:21:59

$(0,5) \ \ \ \ \ \ \ =$ minimum
$(1,8) \ \ \ \ \ \ \ =$ maximum
$(1/2,13/2) =$ non-stationary inflection point

Explanation:

We have:

$f(x) = 5 + 9x^2-6x^3$

We can see the critical point via a graph:

graph{5 + 9x^2-6x^3 [-6, 6, -2, 14]}

We can examine the critical points using calculus:

Differentiating wrt $x$ we get:

$f'(x) = 18x - 18x^2$

At a critical point we have $f'(x) = 0$

$f'(x) = 0 => 18x - 18x^2 = 0$

$:. 18x(1-x) = 0 => x = 0,1$

And, now we have the $x$ -coordinates, we can determine the nature of the turning points (or critical points) by using the second derivative test. Differentiating a second time, we get:

$f''(x) = 18 - 36x$

When:

$x= 0 => f''(0) = 18-0 \ \ gt 0 =>$ minimum
$x= 1 => f''(1) = 18-36 lt 0 =>$ maximum

Also note we have an inflection point if $f''(x)=0$

$f''(x) = 0 => 18-36x = 0 => x=1/2$

Now we have the $x$ -coordinate of the critical points let us find the associated $y$ -coordinate:

$x= 0 \ => f(0) \ \ \ \ \ = 5 + 0-0 = 5$
$x= 1 \ => f(1) \ \ \ \ \ = 5 + 9-6 = 8$
$x= 1/2 => f(1/2) = 5 + 9(1/4)-6(1/8) = 13/2$

Hence, in summary

$(0,5) \ \ \ \ \ \ \ =$ minimum
$(1,8) \ \ \ \ \ \ \ =$ maximum
$(1/2,13/2) =$ non-stationary inflection point

Which is consistent with what we see graphically

Answer 2 · 2017-08-04T17:35:58

The local maximum is $=(1,8)$ and the local minimum is $=(0,5)$
The point of inflection is $=(1/2,13/2)$

Explanation:

Our function is

$f(x)=5+9x^2-6x^3$

The first derivative is

$f'(x)=18x-18x^2$

The critical points are when

$f'(x)=0$

$18x-18x^2=18x(1-x)$

$18x(1-x)=0$

Therefore, $x=0$ and $x=1$

We can build a variation chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $0$ $color(white)(aaaaaaa)$ $1$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x$ $color(white)(aaaaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $1-x$ $color(white)(aaaaa)$ $+$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $-$

$color(white)(aaaa)$ $f'(x)$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $-$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaa)$ $↘$ $color(white)(aaaa)$ $↗$ $color(white)(aaaa)$ $↘$

Now, we calculate the second derivative

$f''(x)=18-36x$

The point of inflection is when $f''(x)=0$

That is,

$18-36x=0$ , $=>$ , $x=18/36=1/2$

We build a variation chart with the second derivative

$color(white)(aaaa)$ $Interval$ $color(white)(aaaa)$ $(-oo,1/2)$ $color(white)(aaaa)$ $(1/2,+oo)$

$color(white)(aaaa)$ $sign f''(x)$ $color(white)(aaaaaa)$ $+$ $color(white)(aaaaaaaaaaaa)$ $-$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaaaaaaa)$ $uu$ $color(white)(aaaaaaaaaaaa)$ $nn$

graph{5+9x^2-6x^3 [-15.35, 16.69, -3.14, 12.88]}

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How do you find the local maximum and minimum values of $f (x) = 5 + 9 x^{2} - 6 x^{3}$ $f(x) = 5 + 9x^2 − 6x^3$ using both the First and Second Derivative Tests?

2 Answers

Explanation:

Explanation:

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Science

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Humanities

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How do you find the local maximum and minimum values of f(x) = 5 + 9x^2 − 6x^3f(x)=5+9x2−6x3f(x) = 5 + 9x^2 − 6x^3 using both the First and Second Derivative Tests?

2 Answers

Explanation:

Explanation:

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How do you find the local maximum and minimum values of $f (x) = 5 + 9 x^{2} - 6 x^{3}$ $f(x) = 5 + 9x^2 − 6x^3$ using both the First and Second Derivative Tests?