How do you use the Mean Value Theorem to solve #f(x)=x - sqrtx# on the closed interval [0,4]?

Question

William Abdalla · Accepted Answer

I assume in your question that you desire to find a value #c# between 0 and 4 so that #f'(c)=\frac{f(4)-f(0)}{4-0}=\frac{2-0}{4-0}=\frac{1}{2}#. Since #f'(c)=1-\frac{1}{2}c^{-1/2}#, that means you want to solve the equation #1-\frac{1}{2}c^{-1/2}=\frac{1}{2}# for #c#. This equation can be rearranged to say that #\frac{1}{2c^{1/2}}=\frac{1}{2}# so that #c^{1/2}=1#. This tells you that #c=1#.

A comment about what we are really doing here: we are not really "using" the Mean Value Theorem to solve anything. Rather, we are solving an algebra equation related to the instantaneous rate of change #f'(c)# and the average rate of change #\frac{f(4)-f(0)}{4-0}# to find the value of #c# that is guaranteed to exist by the Mean Value Theorem over the interval #[0,4]# (a value of #c# where the instantaneous rate of change equals the average rate of change). In effect, we are just illustrating the truth of the theorem, not using it.

Here's a picture of the situation in this problem. Note that the two dashed lines have equal slopes:

Axel Alvarez · Answer

undefined To solve $ f(x) = x - \sqrt{x} $ on the closed interval $[0,4]$ using the Mean Value Theorem:

1. Verify that the function $ f(x) $ is continuous on the closed interval $[0,4]$ and differentiable on the open interval $(0,4)$.
2. Calculate the derivative of $ f(x) $, denoted as $ f'(x) $.
3. Find the average rate of change of $ f(x) $ over the interval $[0,4]$ by evaluating $ \frac{f(4) - f(0)}{4 - 0} $.
4. Apply the Mean Value Theorem, which states that there exists at least one $ c $ in the open interval $(0,4)$ such that $ f'(c) = \frac{f(4) - f(0)}{4 - 0} $.
5. Solve for $ c $ by setting $ f'(c) $ equal to the average rate of change found in step 3.
6. Once you find the value of $ c $, evaluate $ f(c) $ to find the point on the curve where the tangent line is parallel to the secant line.

By following these steps, you can use the Mean Value Theorem to solve $ f(x) = x - \sqrt{x} $ on the closed interval $[0,4]$.