Show that f(x) = [x3 + 1/x3] is decreasing on ]-1, 1[.

1.	Show that f(x) = [x3 + 1/x3] is decreasing on ]-1, 1[.
Answer» It is given that f(x) = [x³ + 1/x³] By differentiating w.r.t. x f’(x) = 3x² – 3x^-4 By taking 3 as common f’(x) = 3[x² – 1/x⁴] Taking LCM f’(x) = 3[(x⁶ – 1)/x⁴] = 3[(x²)³ – 1]/x⁴ On further calculation f’(x) = [3(x² – 1) (x⁴ + x² + 1)]/x⁴ So we get f’(x) = [3(x – 1) (x + 1) (x⁴ + x² + 1)]/x⁴ < 0 for x ∈ (- 1, 1). Therefore, f(x) is decreasing function on ]-1, 1[.

Answer»

It is given that

f(x) = [x³ + 1/x³]

By differentiating w.r.t. x

f’(x) = 3x² – 3x^-4

By taking 3 as common

f’(x) = 3[x² – 1/x⁴]

Taking LCM

f’(x) = 3[(x⁶ – 1)/x⁴] = 3[(x²)³ – 1]/x⁴

On further calculation

f’(x) = [3(x² – 1) (x⁴ + x² + 1)]/x⁴

So we get

f’(x) = [3(x – 1) (x + 1) (x⁴ + x² + 1)]/x⁴ < 0 for x ∈ (- 1, 1).

Therefore, f(x) is decreasing function on ]-1, 1[.

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