Answer:
A
Step-by-step explanation:
We are given a table and the function:
[tex]\displaystyle f(x)=\int_{-2}^{x}g(t)\, dt[/tex]
And:
[tex]h(x)=2x+\sin(x)[/tex]
We want to determine the value of x for which h(x) = f'(2).
First, by the Fundamental Theorem of Calculus:
[tex]f'(x)=g(x)[/tex]
Then by the table:
[tex]f'(2)=g(2)=-2[/tex]
Therefore:
[tex]h(x)=2x+\sin(x)=-2[/tex]
Use graphing technology*:
[tex]x\approx -0.684[/tex]
Our answer is A.
*Perhaps there is a way to solve this manually, though I'm not certain.
