Question

For y=2 - 3/x - 3/x^2, use algebra and calculus to find intervals where the graphs is increasing and decreasing and the x-values of any local extreme...

 

 

Answer 1

The function is y = 2 - 3/x - 3/x².

Differentiate with respect to x.

y' = 3/x² + 6/x³ = (3x+6)/x³.

Again differentiate with respect to x.

y'' = -6/x³ - 18/x⁴ = (-6x -18)/x⁴.

Find the critical numbers at f'(x) = 0 or f'(x) does not exist.

The function f'(x) = 0 when x = -2 and The function f'(x) does not exist when x = 0.

The critical numbers are -2 and 0 and the test intervals are (-∞, -2), (-2, 0) and (0, ∞).

Interval   Test Value               Sign of f'(x)                         Conclusion

(-∞, -2)       x = -3       y'(-3) = [3(-3)+6]/(-3)³ = 1/9 > 0      Increasing

(-2, 0)         x = -1      y'(-1) = [3(-1)+6]/(-1)³ = -3 < 0         Decreasing

(0, ∞)          x = 1       y'(1) = [3(1)+6]/(1)³ = 9 > 0              Increasing

Find Relative Extrema :

The First Derivative Test :

Let f be a differential function with f(c) = 0 then

• If f ' (x) changes from positive (increasing) to negative (Decreasing), f has a relative (local) maximum at c.
• If f ' (x) changes from negative (Decreasing) to positive (increasing), f has a relative (local) minimum at c.

The function f(x) has local maximum at x = - 2, because f'(x) changes from positive to negative around − 2.

y(-2) = 2 - 3/(-2) - 3/(-2)² = 2 + 3/2 - 3/4 = (8 + 6 - 3)/4 = 11/4.

Since x = 0 is not domain of the function y = 2 - 3/x - 3/x².

The relative maximum point (- 2, 11/4).