Question 2.10.3: An Illustration of the Mean Value Theorem Find a value of c ...

An Illustration of the Mean Value Theorem

Find a value of c satisfying the conclusion of the Mean Value Theorem for

f(x)=x^3-x^2-x+1

on the interval [0, 2].

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

Notice that f is continuous on [0, 2] and differentiable on (0, 2). The Mean
Value Theorem then says that there is a number c in (0, 2) for which

f^{\prime}(c)=\frac{f(2)-f(0)}{2-0}=\frac{3-1}{2-0}=1 .

To find this number c, we set

f^{\prime}(c)=3 c^2-2 c-1=1

or 3 c^2-2 c-2=0.

From the quadratic formula, we get c=\frac{1 \pm \sqrt{7}}{3}. In this case, only one of these,

c=\frac{1+\sqrt{7}}{3},is in the interval (0, 2). In Figure 2.53, we show the graphs of y = f (x),

the secant line joining the endpoints of the portion of the curve on the interval [0, 2] and the tangent line at x=\frac{1+\sqrt{7}}{3}.

18

Related Answered Questions

Question: 2.2.7

Verified Answer:

The graph (see Figure 2.18) indicates a sharp corn...
Question: 2.10.5

Verified Answer:

First, note that f (x) = sin x is continuous and d...
Question: 2.10.4

Verified Answer:

We first write down (from our experience with deri...
Question: 2.10.2

Verified Answer:

Figure 2.50 makes the result seem reasonable, but ...
Question: 2.9.2

Verified Answer:

Recall from (9.2) that y=\sinh ^{-1} x \qua...
Question: 2.9.1

Verified Answer:

From the chain rule, we have f^{\prime}(x)=...