Pre-calculus

posted by on .

Use a graphing utility to approximate any relative minimum or maximum values of the function g(x)=x sqrt4-x Not getting this one, either. Thanks.

• typo? - ,

do you mean what you wrote or
x sqrt (4-x) ????

• Pre-calculus - ,

That'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 - ,

if 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 - ,

Here 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 - ,

didn'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 - ,

:)

• Pre-calculus - ,

Why is there a .5 in your problem? And yes, it's x sqrt(4-x)

• ^.5??? - ,

Is ^.5 supposed to be square root? Never heard it put that way before.

• Pre-calculus - ,

easy way to write square root - to the 1/2 power

• Pre-calculus - ,

especially when doing derivatives
d/dx x^(1/2) = (1/2) x^-(1/2)
or
d/dx x^.5 = .5 x^-.5

• Pre-calculus - ,

Thanks!