f (x ) = cos^( 2) ( x ) + sin^( 4) ( x ) R_( f) = ? −−−solution−−− y = cos^( 2) (x ) + sin^( 2) (x) .( 1−cos^( 2) (x)) = 1 − (1/4) sin^( 2) ( 2x) we know that : 0≤ sin^( 2) ( αx) ≤1 therefore −1≤− sin^( 2) (2x) ≤ 0 1−(1/4) ≤ 1− (1/4) sin^( 2) (2x) ≤1 R


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Question Number 161406 by mnjuly1970 last updated on 17/Dec/21

$f (x) = {c o s}^{2} (x) + {s i n}^{4} (x)$ $R_{f} = ?$ $- - - s o l u t i o n - - -$ $y = {c o s}^{2} (x) + {s i n}^{2} (x) . (1 - {c o s}^{2} (x))$ $= 1 - \frac{1}{4} {s i n}^{2} (2 x)$ $w e k n o w t h a t :$ $0 ⩽ {s i n}^{2} (α x) ⩽ 1$ $t h e r e f o r e$ $- 1 ⩽ - {s i n}^{2} (2 x) ⩽ 0$ $1 - \frac{1}{4} ⩽ 1 - \frac{1}{4} {s i n}^{2} (2 x) ⩽ 1$ $R_{f} = [\frac{3}{4}, 1] ◂ ★ ▸$
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