11.18 f(x) = 3x + l.
f'(x) = 3. Hence, f(x) is increasing everywhere.
In Problems 11.18 to 11.26, determine where the function/is increasing and where it is decreasing.
11.17
Prove that, if f'(x)>0 for all x in the open interval (a, b), then f(x) is an increasing function on (a, b).
Assume a some c between u and v, f'(c) = [f(v) - /(«)] /(v - u). Hence, f(v) - /(«) = f'(c)(v - u). Since u v-u>0. By hypothesis, /'(c)>0. Hence, f(v) -/(«)>0, and /(u) >/(«). Thus,/(A:) is increasing
in (a, ft).
11.16
Since x-4 is differentiable and nonzero on [0, 2], so is/(*).
Setting
The value
lies between 0 and 2.
Both of these values lie in [-3,4].
on that interval.
f(x) is differentiable on
since
Setting
we obtain
11.15
we obtain
lies between 1 and 3.
Setting
The value
is differentiable and nonzero on [1,3], f(x) is differentiable on [1,3].
Since
we obtain
11.14
/(*) is continuous for x>0 and differentiable for x>0. Thus, the mean value theorem is applicable.
we find
Setting
11.13 f(x) = x
3 '
4
on [0,16].
11.12 f(x) = 3x
2 - 5x + 1 on [2, 5].
/'(*) = 6*— 5, and the mean value theorem applies. Setting
which lies between 2 and 5.
find
we
70
CHAPTER 11
11.10
State the mean value theorem.
If fix) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there is a
number c in (a, b) such that
In Problems 11.11 to 11.16, determine whether the hypotheses of the mean value theorem hold for the function
f(x) on the given interval, and, if they do, find a value c satisfying the conclusion of the theorem.
11.11
f(x) = 2x + 3 on [1,4].
f'(x) = 2. Hence, the mean value theorem applies. Note that
Thus, we can
take c to be any point in (1,4).
which lies between 0 and 16.
on
on
on