Find the intervals in which the function f given by f(x)=x^3+1/x^3 ,x≠0 is (i) increasing (ii) decreasing
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HELLO DEAR,
GIVEN:- f(x) = x³ + 1/x³
( 1 ) . now, f'(x) = dy/dx(x³ + x^{-3})
=
= (3x² - 3/x⁴)
f'(x) = 3(x² - 1/x⁴)
( 2 ). now put f'(x) = 0
3(x² - 1/x⁴) = 0
x² - 1/x⁴ = 0
(x^6 - 1)/x⁴ = 0
x^6 = 1
x = ±1
HENCE, x = 1 , x = -1
thus value of x = - 1 & 1
i.e. f(x) is strictly increasing on ( -∞ , - 1) & (1 , ∞)
and strictly deacreasing on (- 1 , 1)
I HOPE ITS HELP YOU DEAR,
THANKS
GIVEN:- f(x) = x³ + 1/x³
( 1 ) . now, f'(x) = dy/dx(x³ + x^{-3})
=
= (3x² - 3/x⁴)
f'(x) = 3(x² - 1/x⁴)
( 2 ). now put f'(x) = 0
3(x² - 1/x⁴) = 0
x² - 1/x⁴ = 0
(x^6 - 1)/x⁴ = 0
x^6 = 1
x = ±1
HENCE, x = 1 , x = -1
thus value of x = - 1 & 1
i.e. f(x) is strictly increasing on ( -∞ , - 1) & (1 , ∞)
and strictly deacreasing on (- 1 , 1)
I HOPE ITS HELP YOU DEAR,
THANKS
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