This is the graph of $f$.

Let $g(x)=\int_{0}^{x} f(t) d t$.
What is an appropriate calculus-based justification for the fact that $g$ is concave up?

Choose 1 answer:
(A) $f$ is positive.
(B) $f$ is increasing.
(C) $f$ is concave up.

Question

This is the graph of $f$.

Let $g(x)=\int_{0}^{x} f(t) d t$.
What is an appropriate calculus-based justification for the fact that $g$ is concave up?

Choose 1 answer:
(A) $f$ is positive.
(B) $f$ is increasing.
(C) $f$ is concave up.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We are given the function $g(x) = \int_{0}^{x} f(t) \, dt$ and need to determine why $g$ is concave up. ‖ ‖ step 2 ⋮ To determine the concavity of $g(x)$, we need to consider the second derivative of $g(x)$. ‖ ‖ step 3 ⋮ By the Fundamental Theorem of Calculus, the first derivative of $g(x)$ is $g'(x) = f(x)$. ‖ ‖ step 4 ⋮ The second derivative of $g(x)$ is $g''(x) = f'(x)$. ‖ ‖ step 5 ⋮ For $g(x)$ to be concave up, $g''(x)$ must be positive. This means $f'(x) > 0$. ‖ ‖ step 6 ⋮ Since the graph of $f$ is increasing, $f'(x) > 0$. Therefore, $g(x)$ is concave up because $f$ is increasing. ‖ ∻Answer∻ ⚹ (B) $f$ is increasing. ⚹ ∻Key Concept∻ ⚹Concavity of an integral function⚹ ∻Explanation∻ ⚹The concavity of the integral function $g(x)$ is determined by the sign of the second derivative $g''(x)$. Since $g''(x) = f'(x)$ and $f$ is increasing, $f'(x) > 0$, making $g(x)$ concave up.⚹◊What is the relationship between the concavity of $g(x)$ and the concavity of $f(x)$?⍭ Generate me a similar question◊

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