If f(x) is the function given by f(x) = e^{3x} + 1, at what value of x is the slope of the tangent line to f(x) equal to 2? I thought the derivative of e^{3x} would be 3e^{3x} because of the chain rule, but it doesn't appear to be correct. I know that e^{x} is just e^{x}, so is e^{3x} just e^{3x}?
The region bounded by y = e^x, y = 1, and x = 2 s rotated about the x-axis. The volume of the solid generated is given by the integral_____________. So this is volume problem and a disc/ washer method formula should work best. I was going to set the problem as: pi(antider)from 0-2 of (e^x-1)^2(1-1)^2 but I don't believe I will come out with 1-1 and 0 as the volume.
You are rotating around the x-axis. At a given value of x what is the outer and inner radius of the washer?
Hi sonofjohn! (have a pi: π ) I don't actually understand all of your (e^x-1)^2(1-1)^2 … but you have a π(e^{x} - 1)^{2}, which is not the area of anything, is it?
No, at say x=1 what is the inner and outer radius of the washer? What about at a general value of x? 0 and 2 are fine for the limits on the x integration. Now you just want to find the area of the washer. Correctly, this time. It's not pi*(e^x-1)^2.
Ok so I should find the area of the washer. The area of the washer should be defined as e^x - 1 Also since I am using the dish washer formula, I should square both parts of the integration thus yielding: pi(anitderivative)(e^2x -1)
The 3rd problem on this page is difficult for me to understand. I don't understand what the x stands for when they are talking about the bounds in terms of integration. Do they mean the x-axis on the graph or possible x = 2 where at the x intercept?
x is a point between 0 and 4. Any one. What you might notice is that G(x) is the area under the curve from 0 to x. H(x) is NEGATIVE of the area under the curve from x to 2 (because the integral is from 2 to x instead of from x to 2). Might this tell you something about G(x)-H(x)?
So then the answer must be g(x) = h(x) - 2, because h(x) is positive and g(x) is negative. g(x) is also always going to be two less, because it goes from any point to 0, and h(x) only goes to 2.
I can't really agree with you there, sonofjohn. Try this. What are G(1) and H(1)? Work them out from the picture.
I see now that (d) cannot work. Subtracting 2 everytime from h will not yield an equal integral. Now I would like to say that g(x) = h(x+2) would work, but it doesn't seem plausible past h(1). Could G'(x) = H'(x+2) work? I don't even understand what it means.
You didn't answer my last question. What are G(1) and H(1)? That should let you eliminate some possibilities.