What are the absolute extrema of $f (x) = x \sqrt{25 - x^{2}} \in [- 4, 5]$ $f(x)= xsqrt(25-x^2) in [-4,5]$ ?

Question

What are the absolute extrema of $f (x) = x \sqrt{25 - x^{2}} \in [- 4, 5]$ $f(x)= xsqrt(25-x^2) in [-4,5]$ ?

1 Answer

Impact of this question

2512 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-05T16:17:04

The absolute minimum is $- \frac{25}{2}$ $-25/2$ (at $x = - \sqrt{\frac{25}{2}}$ $x=-sqrt(25/2)$ ). The absolute maximum is $\frac{25}{2}$ $25/2$ (at $x = \sqrt{\frac{25}{2}}$ $x=sqrt(25/2)$ ).

Explanation:

$f (- 4) = - 12$ $f(-4) = -12$ and $f (5) = 0$ $f(5)=0$

$f'(x) = sqrt(25-x^2)+x/(cancel(2)sqrt(25-x^2))*-cancel(2)x$

$= (25-x^2-x^2)/sqrt(25-x^2) = (25-2x^2)/sqrt(25-x^2)$

The critical numbers of $f$ are $x=+-sqrt(25/2)$ Both of these are in $[-4,5]$ ..

$f(-sqrt(25/2)) = -sqrt(25/2)sqrt(25-25/2)$

$= -sqrt(25/2)sqrt(25/2) = -25/2$

By symmetry ( $f$ is odd), $f(sqrt(25/2)) = 25/2$

Summary:
$f(-4) = -12$
$f(-sqrt(25/2)) = -25/2$
$f(sqrt(25/2)) = 25/2$
$f(5)=0$

The absolute minimum is $-25/2$ (at $x=-sqrt(25/2)$ ).
The absolute maximum is $25/2$ (at $x=sqrt(25/2)$ ).

Science

Math

Humanities

... and beyond

What are the absolute extrema of $f (x) = x \sqrt{25 - x^{2}} \in [- 4, 5]$ $f(x)= xsqrt(25-x^2) in [-4,5]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What are the absolute extrema of f(x)= xsqrt(25-x^2) in [-4,5]f(x)=x√25−x2∈[−4,5] f(x)= xsqrt(25-x^2) in [-4,5]?

1 Answer

Explanation:

Related questions

Impact of this question

What are the absolute extrema of $f (x) = x \sqrt{25 - x^{2}} \in [- 4, 5]$ $f(x)= xsqrt(25-x^2) in [-4,5]$ ?