Question

Given: y=2x^2/x^2-1
Find:
1. the X and y intercepts
2. the vertical, horizontal and oblique asymptotes
3. local maximum and the local minimum.
4. limit as x goes to positive infinity.

Answer 1

Given function , y = 2x²/(x²-1)

1. For x-intercept , put y = 0

So 2x²/(x²-1) = 0

=> 2x² = 0

=> x = 0

So x-intercept is (0,0)

For y-intercept , put x = 0

So y = 0/(0-1) = 0

=> y = 0

So y-intercept is (0,0)

2. For vertical asymptotes , x² - 1 =0

=> x² = 1

=> x = ±1

So vertical asymptotes are x = 1, x = -1

For horizontal asymptote,

y = $\frac { coefficient \,of\, x²\, of\, numerator} { coefficient\, of\, x² \,of \,denominator}$ = 2

So horizontal asymptote is y = 2

No oblique asymptote is there as degree of numerator is not higher than that of denominator

3. f'(x) = $\frac { 4x(x²-1)-2x².2x} {(x²-1)²}$

= $\frac { -4x} {(x²-1)²}$

f''(x) = $\frac { -4(x²-1)²+4x.2(x²-1).2x} {(x²-1)⁴}$

= $\frac { 12x²+4} {(x²-1)³}$

For maximum or minimum, f'(x) = 0

=> $\frac { -4x} {(x²-1)²}$ = 0

=> x = 0

Now f''(0) = (12X0+4)/(0-1)³= -4

Since f''(0) < 0, f(x) has local maximum at x = 0 and local maximum value is f(0) = 2X0/(0-1) = 0

The function has no local minimum

4. $\lim_{x → +∞} \frac {2x²} {x²-1}$

= $\lim_{x → +∞} \frac {2} {1-1/x²}$

= 2 since 1/x →0 as x →+∞