So let's just remind ourselves what it means for c to be the number that satisfies the mean value theorem for f. This means that over this interval, c is a point, x equals c where the slope of the tangent line at x equals c so I could write f prime of c, so that is the slope of the tangent line when x is equal to c, this is equal to the slope of the secant line that connects these two points so this is going to be equal to, see the slope of the secant line that connects the points three f of three and one f of one so it's going to be f of three minus f of one over three minus one and if you wanted to think about what this means visually it would look something like this. So this is our x axis and this is one, two actually let me spread it out a little bit more, one, two and three and so you have one comma f of one right over there, so that is at the point one comma f of one and we could evaluate that actually what, that's one comma one, right? So that's going to be the point one comma one and then you have the point three comma, let's see, you're going to have four times three is twelve minus three is nine, so it's going to be three comma three three comma three, so maybe it's right over there three comma three and it might look, the curve might look something like this, so it might look something like that. So if you think about the slope of the line that connects these two points, so this line that connects those two points all the mean value theorem, I'm going to do that in a different color, all the mean value theorem tells us is that there's a point between one and three where the slope of the tangent line has the exact same slope, so if I were to eyeball it, it looks like it's right around there although we are actually going to solve for it, so some point where the slope of the tangent line is equal to the slope of the line that connects these two end points and their corresponding function values, so that is c, that would be c right over there. So really we just have to solve this, so let's first just find out what f prime of x is and then we could substitute a c in there and then we can evaluate this on the right hand side. So I'm going to rewrite f of x f of x is equal to and I'm going to write it as four x to the minus three to the one half power, it makes it a little bit more obvious that we can apply the power rule and the chain rule here, so f prime of x is going to be the derivative of four x minus three to the one half with respect to four x minus three, so that is going to be one half times four x minus three to the negative one half and then we're going to multiply that times the derivative of four x minus three with respect to x, well, derivative of four x with respect to x is just four and the derivative of negative three with respect word x well that's just going to be zero, so the derivative of four x minus three is four, so times four. So f prime of x, f prime of x is equal to four times 1 1⁄2, which is two over the square root of four x minus three.