Question

Find the intervals in which the function f given by f(x)=x3+$\frac{1}{x^3}$,x≠0 is (i) increasing (ii) decreasing

Answer 1

f(x)=$x^3+\frac{1}{x^3}$

f'(x)=3x2-3/x'=3$\times$6-$\frac{3}{x^4}$

Then, f'(x)=0=3$\times$6-3=0=x6=1=x±1

Now, the points x = 1 and x = −1 divide the real line into three disjoint intervals

i.e.,(-∞,-1),(-1,1), and (1,∞).

In intervals (-∞,-1) and (1,∞) i.e., when x < −1 and x > 1, f'(x)>0.

Thus, when x < −1 and x > 1, f is increasing.

In interval (−1, 1) i.e., when −1 < x < 1, f'(x)<0.

Thus, when −1 < x < 1, f is decreasing.