`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.
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.
` ` a) Vertical asymptotes occur where the denominator equal 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.
b) Find the derivative:
Set the numerator equal to 0 (where there are critical points):
Plug in the endpoints and critical values into the original equation
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
Solve for the numerator
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.