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.