uppose $$ f^{\prime}(x)

Question

uppose $$ f^{\prime}(x)<2 $$ for all $$ x \geq 5 $$, and $$ f(5)=8 $$. Show that $$ f(9)<16 $$. which of the following formula holds true? A. $$ f^{\prime}(c)=\frac{b-a}{f(b)-f(a)} $$ B. $$ f^{\prime}(c)=\frac{f(b)+f(a)}{b+a} $$ C. $$ f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} $$ D. $$ f^{\prime}(c)=\frac{f(b)-b}{f(a)-a} $$ Let $$ \mathrm{a}=5 $$ and $$ \mathrm{b}=9 $$. Assume then that $$ f(a)= $$ $$ \square $$ and $$ f(b)= $$ $$ \square $$

UpStudy AI · Accepted Answer

We start by applying the Mean Value Theorem (MVT). For a differentiable function $$ f $$ on the interval $$[a, b]$$, the MVT guarantees that there exists some $$ c $$ in $$(a,b)$$ such that

$$
f'(c) = \frac{f(b) - f(a)}{b - a}.
$$

Given $$ a = 5 $$ and $$ b = 9 $$ (so that $$ f(a) = f(5) $$ and $$ f(b) = f(9) $$), we have

$$
f'(c) = \frac{f(9) - f(5)}{9 - 5} = \frac{f(9)-8}{4},
$$

since we are given that $$ f(5) = 8 $$.

It is also given that $$ f'(x) < 2 $$ for all $$ x \geq 5 $$, so in particular $$ f'(c) < 2 $$. This yields

$$
\frac{f(9)-8}{4} < 2.
$$

Multiplying both sides by 4, we arrive at

$$
f(9) - 8 < 8.
$$

Thus,

$$
f(9) < 16.
$$

This shows that $$ f(9)<16 $$.

Now, let’s compare the formulas given in the options:

A. $$ f'(c) = \frac{b-a}{f(b)-f(a)} $$  
B. $$ f'(c) = \frac{f(b)+f(a)}{b+a} $$  
C. $$ f'(c) = \frac{f(b)-f(a)}{b-a} $$  
D. $$ f'(c) = \frac{f(b)-b}{f(a)-a} $$

From the Mean Value Theorem, we recognize that the correct formula is

$$
f'(c) = \frac{f(b)-f(a)}{b-a},
$$

which corresponds to Option C.

Finally, since we have set $$ a = 5 $$ and $$ b = 9 $$, we fill in the corresponding values:

- $$ f(a) = f(5) = 8 $$
- $$ f(b) = f(9) $$ (with the conclusion $$ f(9) < 16 $$)

Thus, the correct option is C, and we have:

$$
\boxed{f(5)=8, \quad f(9) < 16.}
$$
$$ f(9) < 16 $$, and the correct formula is Option C: $$ f'(c) = \frac{f(b)-f(a)}{b-a} $$.

uppose for all , and . Show that . which of the following formula holds true?
A.
B.
C.
D.

Let and .
Assume then that and

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