The maximum of [tex]f(x)[/tex] occurs at the value of [tex]x[/tex] for which [tex]f'(x)=0[/tex] and the sign of the derivative is positive to the left of [tex]x[/tex] and negative to the right. According to the graph of [tex]f'(x)[/tex], this is the case for [tex]x=6[/tex]: you have [tex]f'(6)=0[/tex], and [tex]f'(x)>0[/tex] for [tex]x[/tex] immediately to the left of 6, and negative immediately to the right.
By the fundamental theorem of calculus, you have
[tex]\displaystyle\int_2^6f'(x)\,\mathrm dx=f(6)-f(2)[/tex]
which means
[tex]\displaystyle f(6)=10+\int_2^6f'(x)\,\mathrm dx[/tex]
You can approximate the integral by finding the area under the curve for the interval [tex][2,6][/tex], which can be done by counting the squares. I count 13 full squares right away, and maybe 5 or 6 additional ones from the remaining pieces. Each square has area 5, so the area is approximately 90 or 95.
So, you have
[tex]f(6)\approx10+90=100[/tex]
or
[tex]f(6)\approx10+95=105[/tex]
If A-D are the only options, that leaves D as the answer.
