Video Example Suppose that f(0) = -7 and f'(x) < 8 for all values of x. How large can f(3) possibly be? EXAMPLE 5 SOLUTION We are given that f is differentiable (and therefore continuous) everywhere. In particular, we can apply the Mean Value Theorem on the interval [0, 3] . There exists a number c such that f(3) f(0) f '(c)7 -0 = SO f(3) f(0)7 f '(c) = -7 + 1 f'(c) We are given that f'(x) < 8 for all x, so in particular we know that f'(c) 7 Multiplying both sides of this inequality by 3, we have 3f '(c) 21 SO f(3) = -7 +7 f'(c) -7 1 The largest possible value for f(3) is

Question
Video Example
Suppose that f(0) = -7 and f'(x) < 8 for all values of x. How large can f(3) possibly be?
EXAMPLE 5
SOLUTION
We are given that f is differentiable (and therefore continuous) everywhere. In particular, we can apply the
Mean Value Theorem on the interval [0, 3] . There exists a number c such that
f(3) f(0) f '(c)7
-0
=
SO
f(3) f(0)7
f '(c) = -7 + 1
f'(c)
We are given that f'(x) < 8 for all x, so in particular we know that f'(c) 7
Multiplying both sides of this
inequality by 3, we have 3f '(c) 21
SO
f(3) = -7 +7
f'(c) -7 1
The largest possible value for f(3) is
Expand
Transcribed Image Text

Video Example Suppose that f(0) = -7 and f'(x) < 8 for all values of x. How large can f(3) possibly be? EXAMPLE 5 SOLUTION We are given that f is differentiable (and therefore continuous) everywhere. In particular, we can apply the Mean Value Theorem on the interval [0, 3] . There exists a number c such that f(3) f(0) f '(c)7 -0 = SO f(3) f(0)7 f '(c) = -7 + 1 f'(c) We are given that f'(x) < 8 for all x, so in particular we know that f'(c) 7 Multiplying both sides of this inequality by 3, we have 3f '(c) 21 SO f(3) = -7 +7 f'(c) -7 1 The largest possible value for f(3) is

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