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Let the function f be continuous and differentiable for all x. Suppose you are given that f(−1)=−3, and that f'(x) for all values of x. Use the Mean Value Theorem to determine the largest possible value of f(5).

Let the function f be continuous and differentiable for all x Suppose you are given that f13 and that fx for all values of x Use the Mean Value Theorem to deter class=

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Answer:

Step-by-step explanation:

By MVT,

[tex]f'(c) = \frac{f(5)-f(-1)}{5-1}[/tex]

[tex]4f'(c)= f(5) + 3\rightarrow f(5) = 4f'(c) - 3[/tex]

Moreover, [tex]f'(x)\leq 4~~~~~~ \forall x,[/tex]

[tex]f(5) \leq 16 - 3 =13[/tex]

Hence the largest possible value of f(5) is 13

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