`f(x) = e^(arctan(x))` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of...

`f(x) = e^(arctan(x))` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts (a)-(d) to sketch the graph of `f`.

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Textbook Question

Chapter 4, 4.3 - Problem 52 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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(a) f is defined and infinitely differential everywhere, so it hasn't vertical asymptotes.

When `x->-oo,` `arctan(x)->-pi/2` and `f(x)->e^(-pi/2) approx 0.21.`

When `x->+oo,` `arctan(x)->+pi/2` and `f(x)->e^(pi/2) approx 4.81.`

So f has horisontal asymptotes `y=e^(-pi/2)` and `y=e^(pi/2).`

(b, c) arctan(x) monotonely increases, and `e^x` monotonely increases, so f(x) also monotonely increases on `RR` and has no maximums or minimums.

(d) now we have to compute f''(x). Recall that `(arctan(x))' = 1/(1+x^2).`

`f'(x) = f(x)*1/(1+x^2),`

`f''(x) = f'(x)*1/(1+x^2) + f(x)*((-2x)/(1+x^2)^2) =(f(x))/(1+x^2)*(1-2x).`

f''(x)=0 only at x=1/2.
f''(x) is positive for x<1/2, so f is concave upward on `(-oo, 1/2).`
f''(x) is negative for x>1/2, so f is concave downward on `(1/2, +oo).`
And x=1/2 is the only inflection point of f.

(e) please look at the picture attached

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