28.08.2022 - 21:32

Gleft( x right) = int_2^x {{1 \over {ln\left(t right)}}dt} for x\ge 2. a. Find G'(x). b. Is G(x) increasing or decreasing? c. Is G(x) concave up or down?

Question:

{eq}G\left( x \right) = int_2^x {{1 \over {ln \left( t \right)}}dt} {/eq} for {eq}x \ge 2 {/eq}.

a. Find G'(x).

b. Is G(x) increasing or decreasing?

c. Is G(x) concave up or down?

{eq}\begin{align}G'(x) \&= \frac{d}{dx}(G(x)) \&= \frac{d}{dx}\left(int_2^x \frac{1}{ln(t)} \dtright) \&= \frac{1}{ln(x)}\end{align}{/eq}

Answers (1)
  • Lora
    April 2, 2023 в 17:38
    a. Using the Fundamental Theorem of Calculus, we have: {eq}\begin{\align*}G'(x) &= \frac{d}{dx}(G(x)) \&= \frac{d}{dx}\left(int_2^x \frac{1}{ln(t)} \dtright) \&= \frac{1}{ln(x)}\end{align*}{/eq} b. To determine if G(x) is increasing or decreasing, we need to analyze its derivative, G'(x). Since G'(x) = 1/ln(x) is always positive for x greater than or equal to 2, G(x) is increasing. c. To determine if G(x) is concave up or concave down, we need to analyze its second derivative, G''(x). Using the quotient rule, we have: {eq}begin{align*} G''(x) &= \frac{d}{dx}left\frac{1}{ln(x)}right) \ &= -frac{1}{x^2ln^2(x)} end{align*}{/eq} Since G''(x) is always negative for x greater than or equal to 2, G(x) is concave down.
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