Interval-in-which-given-function-is-decreasing-f-x-2-x-1-2-x-2-2-

Question Number 36957 by rahul 19 last updated on 07/Jun/18

Interval in which given function is

d e c r e a s i n g .

f (x) = (2^{x} - 1) {(2^{x} - 2)}^{2}

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18

y = (2^{x} - 1) {(2^{x} - 2)}^{2}

\frac{d y}{d x} = (2^{x} - 1) \times 2 (2^{x} - 2) 2^{x} l n 2 + {(2^{x} - 2)}^{2} \times 2^{x} \times l n 2

f o r d e c r e a s i n g \frac{d y}{d x} < 0

(t - 1) \times 2 (t - 2) t l n 2 + l n 2 \times) {(t - 2)}^{2} \times t

l n 2 {2 t \times (t^{2} - 3 t + 2) + t (t^{2} - 4 t + 4)}

= l n 2 {2 t^{3} - 6 t^{2} + 4 t + t^{3} - 4 t^{2} + 4 t}

= l n 2 {3 t^{3} - 10 t^{2} + 8 t)

= l n 2 {t (3 t^{2} - 10 t + 8)}

= l n 2 {t (3 t^{2} - 6 t - 4 t + 8)}

= l n 2 [t (3 t (t - 2) - 4 (t - 2) ∣

= l n 2 t (t - 2) (3 t - 4)

c r i t i c a l v a l u e o f t = 0, t = 2, \frac{4}{3}

2^{x} = 0 x \to - \infty

2^{x} = 2 x = 1

2^{x} = \frac{4}{3} x = 0.41 (a p p r o x)

w h e n t > 2 i, e x > 1 \frac{d y}{d x} > 0 i n c r e a s i n g

w h e n 2 > t > \frac{4}{3} 1 > x > 0.41 \frac{d y}{d x} < 0 d e c r e a s i n g

w h e n \frac{4}{3} > t > 0 0.41 > x > - \infty \frac{d y}{d x} > 0 i n c r e a s i n g

w h e n 0 > t x \to - \infty \frac{d y}{d x} < 0 d e c r e a s i n g

Answered by ajfour last updated on 07/Jun/18

l e t u = 2^{x} \Rightarrow \frac{d u}{d x} = 2^{x} \ln 2

y = (u - 1) {(u - 2)}^{2}

\frac{d y}{d x} = [{(u - 2)}^{2} + 2 (u - 1) (u - 2)] (2^{z} \ln 2)

= (2^{x} \ln 2) (u - 2) (3 u - 4)

\frac{d y}{d x} ⩽ 0 \Rightarrow \frac{4}{3} ⩽ 2^{x} ⩽ 2

\frac{2 \ln 2 - \ln 3}{\ln 2} ⩽ x ⩽ 1 .

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