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

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Suppose $$ f^{\prime}(x)<2 $$ for all $$ x \geq 5 $$, and $$ f(5)=8 $$. Show that $$ f(9)<16 $$. Let $$ a=5 $$ and $$ b=9 $$. $$ f^{\prime}(c)=\frac{f(b)-b}{f(a)-a} $$ Assume then that $$ f(a)=8 $$ and $$ f(b)=16 $$. Substitute the values for $$ a, b, f(a) $$, and $$ f(b) $$ into the formula from the Mean Value Theorem and simplify. . $$ f^{\prime}(c)=2 $$ According to the Mean Value Theorem, if $$ f(9)=\square $$, then on the chosen interval, there must exist some point $$ c $$ such that $$ f^{\prime}(c)=\square $$. However, $$ f^{\prime}(x)<\square $$ So, $$ f(9)<\square $$. $$ \square $$ for all $$ x \geq 5 $$.

UpStudy AI · Accepted Answer

1. By the Mean Value Theorem, for the interval $$[5,9]$$ there exists some $$ c $$ in $$(5,9)$$ such that
   $$
   f'(c)=\frac{f(9)-f(5)}{9-5}.
   $$
2. Given $$ f(5)=8 $$, the expression becomes
   $$
   f'(c)=\frac{f(9)-8}{4}.
   $$
3. We are given that $$ f'(x)<2 $$ for all $$ x\geq 5 $$. In particular, for this $$ c $$ we have
   $$
   \frac{f(9)-8}{4} < 2.
   $$
4. Multiplying both sides by 4 (since 4 is positive) yields
   $$
   f(9)-8 < 8.
   $$
5. Adding 8 to both sides gives
   $$
   f(9)<16.
   $$
$$ f(9) < 16 $$.

Suppose for all , and . Show that .
Let and .
Assume then that and .
Substitute the values for , and into the formula from the Mean Value Theorem and simplify. .

According to the Mean Value Theorem, if , then on the chosen interval, there must exist some point such that . However,
So, . for all .

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Suppose for all , and . Show that . Let and . Assume then that and . Substitute the values for , and into the formula from the Mean Value Theorem and simplify. . According to the Mean Value Theorem, if , then on the chosen interval, there must exist some point such that . However, So, . for all .