# local extrema problem?

• Nov 12th 2008, 08:12 PM
Seigi
local extrema problem?
y=x square root of 8-x^2

I did the derivative, -2x^2 square root of 8-x^2/2 square root 8-x^2
but I don't know if it's right.

Find the intervals on which the function is:
a) increasing?
b) decreasing?
c) concave up?
d) concave down?
e) local extreme values?
f) inflection points?
• Nov 12th 2008, 11:05 PM
earboth
Quote:

Originally Posted by Seigi
y=x square root of 8-x^2

I did the derivative, -2x^2 square root of 8-x^2/2 square root 8-x^2
but I don't know if it's right.

Find the intervals on which the function is:
a) increasing?
b) decreasing?
c) concave up?
d) concave down?
e) local extreme values?
f) inflection points?

I assume that your function is:

$f(x)=x \cdot \sqrt{8-x^2}$

Since the radicand must be positive or zero the domain is: $[-\sqrt{8},\ \sqrt{8}]$

To calculate the first derivation you have to use the product rule:

$f'(x)=\sqrt{8-x^2} \cdot 1 + x\cdot \frac12 (8-x^2)^{-\frac12} \cdot (-2x) = \dfrac{8-2x^2}{\sqrt{8-x^2}}$ with $D_{f'}=(-\sqrt{8},\ \sqrt{8})$

To calculate the concavity you need the second derivation. Use quotient rule:

$f''(x)=\dfrac{\sqrt{8-x^2} \cdot (-4x)-(8-2x^2)\cdot \frac12 (8-x^2)^{-\frac12}\cdot (-2x)}{8-x^2} =$ $\dfrac{(8-x^2)\cdot (-4x)+x(8-2x^2)}{(\sqrt{8-x^2})^3}= \dfrac{2x^3-24x}{(\sqrt{8-x^2})^3}$