The second derivative of a function h is given by h^('')(x)=e^(x^(2))-sin(x)". " On which interval is the graph of h concave up? Use a graphing calculator. Choose 1 answer: (A) x 0.395 (C) x 0 (E) All real numbers

Determine Concave Up: To find where the graph of h is concave up, we need to determine where h''(x) is positive.Find h''(x) > 0: h''(x) = e^(x^2) - sin(x). Since e^(x^2) is always positive, we need to find where e^(x^2) > sin(x).Graph y1 and y2: Use a graphing calculator to graph y1 = e^(x^2) and y2 = sin(x) and find where y1 > y2.Identify Intersection Point: After graphing, it looks like the graphs intersect around x = 0.395. For x > 0.395, y1 is above y2, which means h''(x) is positive.

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The second derivative of a function
h is given by

h^('')(x)=e^(x^(2))-sin(x)". "
On which interval is the graph of
h concave up?
Use a graphing calculator.
Choose 1 answer:
(A)
x < 0.395
(B)
x > 0.395
(C)
x < 0
(D)
x > 0
(E) All real numbers

The second derivative of a function $h$ is given by $\newline$ $h^{\prime \prime}(x)=e^{x^{2}}-\sin (x) \text {. }$ $\newline$ On which interval is the graph of $h$ concave up? $\newline$ Use a graphing calculator. $\newline$ Choose $1$ answer: $\newline$ (A) x<0.395 $\newline$ (B) x>0.395 $\newline$ (C) x<0 $\newline$ (D) x>0 $\newline$ (E) All real numbers

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Full solution

Q. The second derivative of a function

h

is given by

\newline

h^{\prime \prime}(x)=e^{x^{2}}-\sin (x) \text {. }

\newline

On which interval is the graph of

h

concave up?

\newline

Use a graphing calculator.

\newline

Choose

1

answer:

\newline

(A)

x<0.395

\newline

(B)

x>0.395

\newline

(C)

x<0

\newline

(D)

x>0

\newline

(E) All real numbers

Determine Concave Up: To find where the graph of $h$ is concave up, we need to determine where $h''(x)$ is positive.
Find h''(x) > 0: $h''(x) = e^{x^2} - \sin(x)$ . Since $e^{x^2}$ is always positive, we need to find where e^{x^2} > \sin(x).
Graph $y_1$ and $y_2$ : Use a graphing calculator to graph $y_1 = e^{x^2}$ and $y_2 = \sin(x)$ and find where y_1 > y_2.
Identify Intersection Point: After graphing, it looks like the graphs intersect around $x = 0.395$ . For x > 0.395, $y_1$ is above $y_2$ , which means $h''(x)$ is positive.