You sure on that domain?
2x-1 = 0 ==> x = 1/2
2x-1 = 6 ==> x = 7/2
since h'(x) = 2f(2x-1), h(x) is a max when f(2x-1) = 0
f(4) = 0, so 2x-1=4 ==> x = 5/2
Let f be the function shown below, with domain the closed interval [0, 6]. Let h(x) = integral from 0 to 2x-1 (f(t))dt.
The graph is a semicircle above the x-axis with r=2 and center (2,0) connected to a semicircle below the x-axis with r=1 and center (5,0).
A. Find h'(x)
B. At what x is h(x) a maximum?
So I determined the domain for h(x) = .5 ≤ x ≤ 1.5 and differentiated h(x) to get h'(x) = 2f (2x-1), but I don't know how to solve the rest. Any help is appreciated :)
4 answers
thank you! sorry the domain was a mistype.
how do i solve for h'(5/2)?
does the equation integral from 0 to x (2-t) / √4-(t-2)^2 work?
how do i solve for h'(5/2)?
does the equation integral from 0 to x (2-t) / √4-(t-2)^2 work?
^ the equation for f
nvm i figured it out. thanks for your help !!