How do I find the numbers #c# that satisfy Rolle's Theorem for #f(x)=sqrt(x)-x/3# on the interval #[0,9]# ?

Question

Paisley Johnson · Accepted Answer

#c=9/4#

Let us look at some details.

By taking the derivative,

#f'(x)=1/{2sqrt{x}}-1/3#

By solving #f'(c)=0#,

#f'(c)=1/{2sqrt{c}}-1/3=0#

by adding #1/3#,

#Rightarrow 1/{2sqrt{c}}=1/3#

by taking the reciprocal,

#Rightarrow 2sqrt{c}=3#

by dividing by 2,

#Rightarrow sqrt{c}=3/2#

by squaring,

#Rightarrow c=9/4#

Aurora Alvarado · Answer

undefined To find the numbers $ c $ that satisfy Rolle's Theorem for the function $ f(x) = \sqrt{x} - \frac{x}{3} $ on the interval $[0,9]$, we need to check if the conditions of Rolle's Theorem are met:

1. $ f(x) $ is continuous on the closed interval $[0,9]$.
2. $ f(x) $ is differentiable on the open interval $(0,9)$.
3. $ f(0) = f(9) $.

Let's evaluate these conditions:

1. $ f(x) = \sqrt{x} - \frac{x}{3} $ is continuous on the closed interval $[0,9]$ because it is a combination of continuous functions.
2. $ f(x) = \sqrt{x} - \frac{x}{3} $ is differentiable on the open interval $(0,9)$ because both $ \sqrt{x} $ and $ \frac{x}{3} $ are differentiable on that interval.
3. $ f(0) = \sqrt{0} - \frac{0}{3} = 0 $ and $ f(9) = \sqrt{9} - \frac{9}{3} = 3 - 3 = 0 $. Therefore, $ f(0) = f(9) $.

Since all conditions of Rolle's Theorem are satisfied, there exists at least one number $ c $ in the open interval $(0,9)$ such that $ f'(c) = 0 $.

To find $ c $, we need to find the derivative of $ f(x) $ and set it equal to zero, then solve for $ x $.

$ f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{3} $

Setting $ f'(x) $ equal to zero:

$ \frac{1}{2\sqrt{x}} - \frac{1}{3} = 0 $

$ \frac{1}{2\sqrt{x}} = \frac{1}{3} $

Solving for $ x $:

$ \sqrt{x} = \frac{3}{2} $

$ x = \left(\frac{3}{2}\right)^2 $

$ x = \frac{9}{4} $

So, $ c = \frac{9}{4} $. Therefore, the number $ c $ that satisfies Rolle's Theorem for the given function on the interval $[0,9]$ is $ \frac{9}{4} $.