# Calculus

posted by .

Find the critical numbers of y=x(4-x)^1/2.
The answer is x=8/3 but I got x=7/2.

• Calculus -

y = x√(4-x)
y' = 1√(4-x) + x * 1/2 * 1/√(4-x) * -1
= √(4-x) - x/(2√(4-x))
= (2(4-x) - x)/(2√(4-x))
= (8-3x)/(2√(4-x))

assuming x≠4, we just have to find where the numerator is zero.

x = 8/3

## Similar Questions

1. ### Calculus

How do you find the critical numbers and the end points of this function: f(x)=2x^3-3x^2-12x+5 i calculated the critical numbers as x=-1 and x=2 is this right?
2. ### Calculus - another question

"Find the absolute minimum value of the function f(x) = x ln(x)". I know that one of the critical values is 1 (domain restriction). Am I right?
3. ### Calculus

Find the critical numbers of the function g(t)=|5t-7| If you find g'(t) you just get 5 so would there be no critical numbers?
4. ### calculus

Consider a differentiable functionf having domain all positive real numbers, and for which it is known that f'(x)=(4-x)^-3 for x.0 a. find the x-coordinate of the critical point f. DEtermine whether the point is a relative maximum …
5. ### math

If y= (x^2+20)/(x-4) Find all critical Numbers. After get f' i got the answer -5/4. What did i do wrong or is that the critical number?
6. ### Calculus

If y= (x^2+20)/(x-4) Find all critical Numbers. After get f' i got the answer -5/4. What did i do wrong or is that the critical number?
7. ### calculus

Please help!! *f(x)= x(x+6)^1/2 find two x intercepts. Then show that f'(x)=0 at some point between the 2 x intercepts. *Use mean value theorem for f(x)= x(x^2-3x+1) for interval [1,4] I FOUND DERIAVITIVE WHICH IS 3X^2-6X+1.THEN?