3, x is equal to 3, is still in the running for the x-coordinate where g has a global absolute maximum. Now what we have to do is figure out any critical points that g has in between. So points that it's either undifferentiable or its derivative is equal to 0. So let's look at its derivative. So g prime of x, we just take the derivative of this business up here, the derivative of 2x is 2, derivative of this definite integral from 0 to x of f of t dt, we did that in Part A, this is just the fundamental theorem of calculus, this is just going to be plus f of x right over there. So it actually turns out that g is differentiable over the entire interval. You give any x value over this interval, we have a value for f of x. f of x isn't differentiable everywhere, but definitely f of x is defined everywhere over the interval.