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March 4, 2015

March 4, 2015

Posted by **Ty** on Monday, November 21, 2011 at 6:00pm.

- typo? -
**Damon**, Monday, November 21, 2011 at 6:10pmdo you mean what you wrote or

x sqrt (4-x) ????

- Pre-calculus -
**Reiny**, Monday, November 21, 2011 at 6:16pmThat's because there is no maximum/minimum

g(x) = √(4-x) = (4-x)^(1/2)

g '(x) = (1/2)(4-x)^(-1/2) (-1)

= -1/(2√(4-x)

This can never be zero

- Pre-calculus -
**Damon**, Monday, November 21, 2011 at 6:18pmif the latter

y = x (4-x)^.5

dy/dx = x (.5)(4-x)^-.5)(-1) + (4-x)^.5

when is that zero

0 = -.5x /(4-x)^.5 + (4-x)^.5

0 = -.5 x + (4-x)

4 = 1.5 x

x = 2.67 for max or min

You should find a max or min around there with your grapher

You can find out with the second derivative but you should see it on your graph anyway

I will find a grapher online

for function

x * sqrt(4-x)

- Pre-calculus -
**Damon**, Monday, November 21, 2011 at 6:19pmHere you go

http://www.coolmath.com/graphit/

put in

x * sqrt(4-x)

and hit enter. You will see the max at 2.67

- oops - Pre-calculus -
**Reiny**, Monday, November 21, 2011 at 6:23pmdidn't see that x in front of √

so by the product rule

I got

g'(x) = (1/2)(3x-8)(4-x)^(-1/2)

which when we set that to zero

x = 8/3

your graphing calculator should show a max point near x = 8/3 , (near about x=3)

- Pre-calculus -
**Damon**, Monday, November 21, 2011 at 6:24pm:)

- Pre-calculus -
**Ty**, Monday, November 21, 2011 at 6:25pmWhy is there a .5 in your problem? And yes, it's x sqrt(4-x)

- ^.5??? -
**Ty**, Monday, November 21, 2011 at 6:35pmIs ^.5 supposed to be square root? Never heard it put that way before.

- Pre-calculus -
**Damon**, Monday, November 21, 2011 at 6:40pmeasy way to write square root - to the 1/2 power

- Pre-calculus -
**Damon**, Monday, November 21, 2011 at 6:41pmespecially when doing derivatives

d/dx x^(1/2) = (1/2) x^-(1/2)

or

d/dx x^.5 = .5 x^-.5

- Pre-calculus -
**Ty**, Monday, November 21, 2011 at 8:33pmThanks!

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