`f(x) = (x^2 - 4)/(x^2 + 4)` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values.

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Textbook Question

Chapter 4, 4.3 - Problem 46 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the vertical asymptotes of the given rational function, hence, you need to find out the zeroes of denominator.

Putting `x^2+4 = 0` yields that there are no real values for x such as `x^2+4 = 0` , hence, there are no vertical asymptotes.

You need to evaluate the horizontal asymtpotes, hence, you need to evaluate the following limits, such that:

`lim_(x->+-oo) (x^2 - 4)/(x^2+4) = 1`

Hence, the horizontal asymptote of the function is y = 1.

b) You need to evaluate the monotony of the function, hence, you need to determine the intervals where the derivative is positive or negative.

`f'(x) = ((x^2-4)/(x^2+4))'`

Using the quotient rule yields:

f'`(x) = ((x^2-4)'(x^2+4) - (x^2-4)(x^2+4)')/((x^2+4)^2)`

f'`(x) = (2x(x^2+4) - 2x(x^2-4))/((x^2+4)^2)`

Factoring out 2x yields:

f'`(x) = 2x(x^2+4 - x^2+ 4)/((x^2+4)^2)`

`f'(x) = (16x)/((x^2+4)^2)`

Putting f'(x) = 0, yields:

`(16x)/((x^2+4)^2) = 0 => 16x = 0 => x = 0`

You need to notice that f'(x)<0, hence the function decreases,  for `x in (-oo,0) ` and f'(x)>0 and the function increases, for `x in (0,+oo).`

c) The maximum and minimum points occurs at the values of x for f'(x) = 0.

From previous point b) yields that f'(x) = 0 for x = 0 and from the monotony of the function yields that (0,-1) is a minimum point.

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scisser | (Level 3) Honors

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` ` a) Vertical asymptotes occur where the denominator equal 0

`x^2+4=0 `

The values of x are imaginary, therefore there are no real vertical asymptotes.

Horizontal asymptotes: If both polynomials are the same degree, divide the coefficients of the highest degree terms.

`y=1 `

b) Find the derivative:

`f'(x)=((x^2+4)(2x)-(x^2-4)(2x))/(x^2+4)^2 `

`f'(x)=(16x)/(x^2+4)^2 `

Set the numerator equal to 0 (where there are critical points):

`16x=0 `

`x=0 `

Plug in the endpoints and critical values into the original equation

`f(-4)=3/5 `

`f(0)=-1 `

`f(4)=3/5 `

This function is decreasing from `(-4,0) ` and increasing from `(0,4) `

Thus, the min is -1 at x=0 and the max is `3/5 ` at both x=-4 and x=4

d) Find the second derivative

`f''(x)=(-16(3x^2-4))/(x^2+4)^3 `

Solve for the numerator

`-16(3x^2-4)=0 `

`x=+-sqrt(4/3)`

For `xlt-sqrt(4/3) f''(x)lt0` so the function is concave down.

For `xgt-sqrt(4/3) f''(x)gt0 ` so the function is concave up

 

For `xltsqrt(4/3) f''(x)gt0` So the function is concave up

For `xgtsqrt(4/3) f''(x)lt0` So the function is concave down.

 

Both `+-sqrt(4/3)` are inflection points.

e) graph

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