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

Question

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

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To apply the Mean Value Theorem for $g(x) = \sqrt{2x - 4}$ on the interval $[2, 10]$, we first need to find $g'(x)$. ‖ ‖ 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.⚹◊What is the Mean Value Theorem and how is it applied to functions like $g(x) = \sqrt{2x-4}$ on the interval $[2,10]$?⍭ Generate me a similar question◊

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