2. (5 points) For each of the functions f and g, find its...

2. (5 points) For each of the functions $f$ and $g$, find its domain, $x$- and $y$-intercepts, limits at the boundary points of the domain and at points of discontinuity, local and global extrema, (maximal) monotonicity intervals, and (maximal) convexity/concavity intervals. Sketch a graph of the function that is consistent with your findings. Justify your answers sufficiently. (a) $f(x) = e^{\frac{x^2+1}{x^2-1}}$ (b) $g(x) = \arctan(\frac{x+1}{x-1})$

Transcript

00:01 Since there's a lot going on, i'm just going to jump right into answering a few of these questions.

00:11 Like on letter a, when they ask you to do g of 2, you have to plug in 2 for that at x.

00:18 So, 2 times 2 is 4.

00:20 And what you really need to do now is that we're doing area under the curve.

00:24 So if i'm looking at from 0 to 4, doing that area, you first have a triangle whose area is 1 half.

00:33 Then you have a semicircle with a radius of 1.

00:38 I should probably write out r equals 1.

00:40 And then from 3 to 4 is another 1 half.

00:45 So adding the halves together, 1 half plus 1 half would be 1.

00:49 And then pi r squared, so pi times 1 squared is 1, divided by 2 because it's only half of a circle.

00:57 And it's below the x -axis, so it should actually be negative.

01:00 I guess i should mention that being below the x -axis is a negative area.

01:05 Pi 1 squared over 2.

01:09 Okay, so looking at the next one, in letter a, they want you to do the derivative of 2.

01:16 Well, the first thing i would point out is that g prime will cancel out the derivative, except it's the chain rule.

01:22 So it's going to be f of 2x times the derivative of 2x, which is 2.

01:27 So what you really have to do when you do g prime of 2 is you need to know what f of 4 is, and then multiply by 2.

01:38 So if you look closely at f of 4, you get an answer of 1, and then multiply by 2 and you get 2.

01:47 So the next one is local mins and maxes for g.

01:54 And it's rooted in this equation.

01:57 So i'm doing letter b, because if you have a local min or a local max, the derivative needs to equal 0.

02:09 And so what we're really looking at is where 2f of 2x will equal 0.

02:15 So that'll only happen as if f of 2x equals 0...

Question

Please give Ace some feedback