How do you determine all values of c that satisfy the mean value theorem on the interval [0,1] for # sqrt (x(1-x))#?

Question

Evelyn Agard · Accepted Answer

#c=1/2# According to the therem, #c# is all x-values that satisfy #f'(x)=(Deltaf(x))/(Deltax)# (within the given interval) #f(x)=sqrt(x(1−x))# #f'(x)=( 1-2x)/(2sqrt(x(1−x)))# #(Deltaf(x))/(Deltax)=[f(1)-f(0)]/[1-0]=0# Equating, #( 1-2x)/(2sqrt(x(1−x)))=0# #1-2x=0# #x=1/2# #:.c=1/2#

Ezra Adams · Answer

undefined To find all values of $ c $ that satisfy the Mean Value Theorem (MVT) for the function $ f(x) = \sqrt{x(1-x)} $ on the interval $[0,1]$, follow these steps:

1. Calculate the derivative of $ f(x) $: $ f'(x) = \frac{1}{2\sqrt{x(1-x)}} - \frac{x - 1}{2\sqrt{x(1-x)}} $.
2. Simplify the derivative: $ f'(x) = \frac{1 - 2x}{2\sqrt{x(1-x)}} $.
3. Apply the MVT: There exists $ c $ in $(0,1)$ such that $ f'(c) = \frac{f(1) - f(0)}{1 - 0} $.
4. Find $ f(1) $ and $ f(0) $: $ f(1) = \sqrt{1(1-1)} = 0 $ and $ f(0) = \sqrt{0(1-0)} = 0 $.
5. Substitute into the MVT equation: $ f'(c) = \frac{0 - 0}{1 - 0} = 0 $.
6. Set the derivative equal to 0 and solve for $ c $: $ \frac{1 - 2c}{2\sqrt{c(1-c)}} = 0 $.
7. Solve for $ c $: $ 1 - 2c = 0 \Rightarrow 2c = 1 \Rightarrow c = \frac{1}{2} $.

Therefore, the only value of $ c $ in the interval $[0,1]$ that satisfies the Mean Value Theorem for $ f(x) = \sqrt{x(1-x)} $ is $ c = \frac{1}{2} $.