4 for all values of x. How large can f(3) possibly be? EXAMPLE 5Suppose that f(0) = -2 and f'(x) We are given that f is differentiable (and therefore continuous) everywhere. In particular, we can apply the SOLUTION Mean Value Theorem on the interval O, 31. There exists a number c such that ro(-0) f(3) (0) f(c) So f(3) f(0) f'(c) f'(c). =-2 + We are given that f'(x) s 4 for all x, so in particular we know that f'(c) Multiplying both sides of this inequality by 3, we have 3f'(c) So f(3) 2 f(c) s -2+ The largest possible value for f(3) is

Question
4 for all values of x. How large can f(3) possibly be?
EXAMPLE 5Suppose that f(0) = -2 and f'(x)
We are given that f is differentiable (and therefore continuous) everywhere. In particular, we can apply the
SOLUTION
Mean Value Theorem on the interval O, 31. There exists a number c such that
ro(-0)
f(3) (0) f(c)
So
f(3) f(0)
f'(c)
f'(c).
=-2 +
We are given that f'(x) s 4 for all x, so in particular we know that f'(c)
Multiplying both sides of this
inequality by 3, we have 3f'(c)
So
f(3) 2
f(c) s -2+
The largest possible value for f(3) is

Image Transcription

4 for all values of x. How large can f(3) possibly be? EXAMPLE 5Suppose that f(0) = -2 and f'(x) We are given that f is differentiable (and therefore continuous) everywhere. In particular, we can apply the SOLUTION Mean Value Theorem on the interval O, 31. There exists a number c such that ro(-0) f(3) (0) f(c) So f(3) f(0) f'(c) f'(c). =-2 + We are given that f'(x) s 4 for all x, so in particular we know that f'(c) Multiplying both sides of this inequality by 3, we have 3f'(c) So f(3) 2 f(c) s -2+ The largest possible value for f(3) is

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