Calculus questions

Question

Hello,

 

I need help to answers these questions.

 

1- Let f(x)=e2x -3x. ? Find an equation for the tangent line to y = f(x) when x = 0. ? Find any critical numbers for f (x) and determine if each produces a local maximum point, a local minimum point, or neither. 

 

2-Let R be the region in the plane bounded by y=1x,y= 1 ,andx=4. x ? Sketch the planar region. ? Find the area of R. ? Find the volume of the solid of revolution obtained by rotating R about the x axis.

Answer 1

For the first question,

 

f = e^2x - 3x means that f' = 2e^2x - 3

 

We want to find the equation of the line that's tangent to the graph of f when x is 0. Let's find the point that line will touch the graph by taking f(0) = e²⁽⁰⁾ - 3(0) = 1 - 0 = 1. So when x is 0, y is 1. The point the tangent will hit the graph is (0,1).

 

To find the slope of the tangent at that point, we recognize that that IS what the derivative is supposed to provide for us. f'(0) IS the slope of the line tangent to f at that point. So, let's find f'(0). f'(0) = 2e²⁽⁰⁾ - 3 = 2(1) - 3 = -1.

 

What do you have now for that line? You have a point on it, and you have its slope. Point slope form is a solid choice:

 

since our point is (0,1) and our m = -1

 

y - 1 = -1(x - 0)               gives us an equation for the tangent line. You can simplify this further: 

 

y - 1 = -x

 

y = -x + 1

 

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Still on the first question, we find critical points of a function where its derivative is 0. So let's take what we got for f' and set it equal to 0:

 

2e^2x - 3 = 0

 

e^2x = 3/2

 

2xlog(e) = log(3/2)

 

2x = log(3/2)

 

x = (1/2)log(3/2)

 

To know whether this is a maximum or a minimum, we can check the sign of the derivative in the interval to the left of this point, and in the interval to the right. 

 

I'm going to estimate (1/2)log(3/2) to be about 0.088. So I'll choose a convenient point on the left and a convenient point on the right. Let's say we pick 0 and 1 because 0.088 is between those. 

 

f'(0) we've already found to be -1. Let's check f'(1) then. 

 

f'(1) = 2e²⁽¹⁾ - 3 = 2e² - 3. Since e itself is in excess of 2, we can safely conclude this is positive without doing anymore work.

 

Thus we have a critical point at about 0.088. The derivative is negative to the left of it (remember that the derivative gives the slope of f) so the slope of f is negative to the left. The derivative is positive to the right. So the slope of f is positive to the right. We don't have any discontinuities in the graph to mess us up, so we can picture the graph of f sloping DOWN to a minimum and then back UP again as we look from left to right. So f must have a minimum at that point. 

 

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The second question,

 

If you're having trouble graphing the region, remember that y = 1 gives you a horizontal line that passes through 1 on the y axis and that x = 4 gives you a vertical line that passes through 4 on the x axis. y = x exactly splices the first quadrant in two and passes through the origin. Two points on y = x are the origin and (1, 1). That's all you need to graph that line. 

 

The region bounded by those lines is a triangle with vertices at (1, 1), (4, 1), and (4, 4). This gives a triangle with a base of 3 and a height of 3. 

 

You can find the area of this without resorting to an integral. 

 

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Picture rotating this triangle about the x axis. What it's going to generate is a truncated cone with a big hole bored down the center that will have a radius of 1. It may be nicer to consider finding the volume of that truncated cone first and then taking away from that the volume of the cylinder of the same height and the radius of 1. 

 

To find the volume of the solid of revolution, imagine having a bunch of little disks with holes in them stacked up in that solid to approximate its volume. How do you find the volume of a shape like that? Treat each of those disks like a cylinder where the height is the change in x and the outer radius is the vertical distance to the line y = x.

 

The volume of one of those little cylinders would be πr²(x₂ - x₁) where r itself is the distance from the x axis to the line y = x. I'll under approximate and pick the left endpoint: x₁. Then r = x₁ and the volume would be πx₁²(x₂ - x₁). 

 

If I split the height of that entire figure up into equal subintervals, then I don't need to consider pairs of x_is the whole time and can write their difference as being Δx because the height of each cylinder will be the same. 

 

 

Then however many subintervals I split [1,4] into, I can approximate the volume of the cone by adding up the volumes of all those little cylinders:

 

∑πx_i²Δx = π∑x_i²Δx = π∫₁⁴ x²dx. = π[(1/3)x³]₁⁴ = (π/3)[4³ - 1³] = (π/3)(63) = 21π.

 

(By the way, that's not supposed to be 1 to the 4th there, those are the bounds of the integral.)

 

However, note that this is the volume of the cone without the hole in it. So we need to take away the volume of the hole. 

 

The volume of the hole is that of a cylinder with a height of 3 and a radius of 1. So 3π. 

 

Then 21π - 3π = 18π. 

 

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There is another way to find the volume of this figure without resorting to integrals. Imagine this time the cone has its tip when you rotate y = x about the x axis. The volume of the whole thing should be that of a cone of height 4 and radius 4, take away the volume of a cone of height 1 and radius 1 (the tip that's missing), take away the volume of the cylindrical hole (which we just found to be 3π above).

 

The volume of the big cone is (1/3)π(4²)(4) = 1/3π4³. 

 

The volume of the little cone is (1/3)π(1²)(1) = (1/3)π.

 

So their difference is (1/3)π(64 - 1) = (1/3)π(63) = 21π, the same thing we got from the integral. Which we hoped would be the case. (It's too easy to make mistakes.) 

 

Take away from this again the volume of the cylindrical hole and we find the volume of this solid of revolution to be 18π. 

 

 

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Apologies for any typos or other errors.

Answer 2

f(x)= e^2x-3x

f'(x) = 2e^2x-3

When x = 0, f(x) = 1 and f'(x) = -1

Assuming the equation for the tangent line y = mx + b

where m is the slope of the line = -1 at x = 0

At x = 0, 1 = 0 + b

Therefore, the equation of the tangent line y = -x + 1