Question

Math

Posted 5 months ago

```
Let $g(x)=\sqrt{2 x-4}$ and let $c$ be the number that satisfies the Mean Value Theorem for $g$ on the interval $2 \leq x \leq 10$.
What is $c$ ?
Choose 1 answer:
(A) 2.25
(B) 3.75
(C) 4
(D) 6
```

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Answer from Sia

Posted 5 months ago

Solution by Steps

step 2

Differentiate $g(x)$ with respect to $x$ to get $g'(x) = \frac{1}{2\sqrt{2x - 4}} \cdot 2 = \frac{1}{\sqrt{2x - 4}}$

step 3

According to the Mean Value Theorem, there exists some $c$ in $(2, 10)$ such that $g'(c) = \frac{g(10) - g(2)}{10 - 2}$

step 4

Calculate $g(10)$ and $g(2)$: $g(10) = \sqrt{2 \cdot 10 - 4} = \sqrt{16} = 4$ and $g(2) = \sqrt{2 \cdot 2 - 4} = 0$

step 5

Substitute $g(10)$ and $g(2)$ into the equation from step 3 to find $g'(c) = \frac{4 - 0}{10 - 2} = \frac{4}{8} = \frac{1}{2}$

step 6

Set $g'(c) = \frac{1}{\sqrt{2c - 4}}$ equal to $\frac{1}{2}$ and solve for $c$: $\frac{1}{\sqrt{2c - 4}} = \frac{1}{2}$

step 7

Squaring both sides of the equation gives $2c - 4 = 4$

step 8

Solve for $c$: $2c = 8$ and $c = 4$

C

Key Concept

Mean Value Theorem

Explanation

The Mean Value Theorem states that for a continuous function on a closed interval and differentiable on the open interval, there exists at least one point $c$ where the instantaneous rate of change (derivative) is equal to the average rate of change over the interval.

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