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

Question

Suppose $$ f^{\prime}(x)<2 $$ for all $$ x \geq 5 $$, and $$ f(5)=8 $$. Show that $$ f(9)<16 $$. The Mean Value Theorem states that if $$ f $$ is continuous on the closed interval $$ [a, b] $$ and differentiable on $$ (a, b) $$, then there is at least one point $$ c $$ in ( $$ a, b $$ ) such that 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 $$ a=\square $$ and $$ b=\square $$.

UpStudy AI · Accepted Answer

We need to show that $$ f(9) < 16 $$ given that $$ f'(x) < 2 $$ for all $$ x \geq 5 $$ and $$ f(5) = 8 $$.

We will apply the Mean Value Theorem (MVT) on the interval $$ [5, 9] $$. According to the MVT, there exists a point $$ c $$ in $$ (5, 9) $$ such that:

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

Here, we can identify $$ a = 5 $$ and $$ b = 9 $$. Thus, we can rewrite the equation as:

$$
f'(c) = \frac{f(9) - 8}{4}
$$
Since we know that $$ f'(x) < 2 $$ for all $$ x \geq 5 $$, it follows that:

$$
f'(c) < 2
$$

Substituting this into the MVT equation gives us:

$$
\frac{f(9) - 8}{4} < 2
$$
Now, we can manipulate the inequality to find $$ f(9) $$:

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

Adding 8 to both sides results in:

$$
f(9) < 16
$$
Thus, we have shown that $$ f(9) < 16 $$ using the Mean Value Theorem. The values for $$ a $$ and $$ b $$ are $$ a = 5 $$ and $$ b = 9 $$.

Suppose for all , and . Show that .
The Mean Value Theorem states that if is continuous on the closed interval and differentiable on , then there is at least one point in ( ) such that
of the following formula holds true?
A.
B.
C.
D.
Let and .

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Suppose for all , and . Show that . The Mean Value Theorem states that if is continuous on the closed interval and differentiable on , then there is at least one point in ( ) such that of the following formula holds true? A. B. C. D. Let and .