106 STEP 4. Review the Knowledge You Need to Score High
Example 3
The pointsP(1, 1) andQ(3, 27) are on the curve f(x)=x^3. Using the Mean Value
Theorem, findc in the interval (1, 3) such that f′(c) is equal to the slope of the
secantPQ.
The slope of secantPQism=27 − 1
3 − 1
=13. Since f(x) is defined for all real numbers,
f(x) is continuous on [1, 3]. Alsof′(x)= 3 x^2 is defined for all real numbers. Thus, f(x)
is differentiable on (1, 3). Therefore, there exists a numbercin (1, 3) such that f′(c)=13.Setf′(c)= 13 ⇒3(c)^2 =13 orc^2 =13
3
c=±√
13
3. Since only
√
13
3
is in the interval(1, 3),c=√
13
3. (See Figure 7.1-5.)
[– 4,4] by [–20,40]
Figure 7.1-5
Example 4
Let f be the function f(x)=(x−1)^2 /^3. Determine if the hypotheses of the Mean Value
Theorem are satisfied on the interval [0, 2], and if so, find all values ofcthat satisfy the
conclusion of the theorem.
Entery 1 =(x−1)^2 /^3. The graphy1 shows that there is a cusp atx=1. Thus, f(x)isnot
differentiable on (0, 2), which implies there may or may not exist acin (0, 2) such that
f′(c)=
f(2)− f(0)
2 − 0. The derivative f′(x)=
2
3
(x−1)−^1 /^3 and
f(2)−f(0)
2 − 0=
1 − 1
2
=0. Set
2
3
(x−1)^1 /^3 = 0 ⇒x=1. Note thatfis not differentiable (a+x=1). Therefore,cdoes not
exist. (See Figure 7.1-6.)[–8,8] by [–4,4]
Figure 7.1-6