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给定两个函数: f ( x ) = x + 3. f (x) = x + 3 f(x)=x+3. g ( x ) = x 2 g (x) = \ dfrac {x} {2} g(x)=2x 找到…的价值 g ( f ( 5 ) ) g (f (5)) g(f(5)).
f ( x ) = x + 3. f (x) = x + 3 f(x)=x+3.
g ( x ) = x 2 g (x) = \ dfrac {x} {2} g(x)=2x
找到…的价值 g ( f ( 5 ) ) g (f (5)) g(f(5)).
对于这两个函数 f ( x ) = x 2 + 5 x , g ( x ) = x + 22 , F (x)=x²+5x, \, g(x)=x+22, f(x)=x2+5x,g(x)=x+22,价值是什么 ( g ∘ f ) ( 9 ) ? (g \circ f)(9)? (g∘f)(9)?
细节和假设
( g ∘ f ) ( x ) (g \circ f)(x) (g∘f)(x)是由 g g g而且 f f f,定义为 ( g ∘ f ) ( x ) = g ( f ( x ) ) (g \circ f)(x)= g(f(x)) (g∘f)(x)=g(f(x)).
让 f f f是这样的函数 ( f ∘ f ) ( x ) = x 2 (f \circ f)(x) = x^2 (f∘f)(x)=x2
价值是什么 f ( f ( f ( f ( f ( f ( 2 ) ) ) ) ) ) ? f (f (f (f (f (f (2)))))) ? f(f(f(f(f(f(2))))))?
考虑这两个函数 f ( x ) = 6 x − 5 , g ( x ) = { − x + 4 如果 x ≥ 1 , 17 如果 x < 1. f (x) = 6 * 5 g (x) ={病例}- x + 4 & \ \开始文本{如果}x \组1、17 \ \ & \文本{如果}x < 1。\{病例}结束 f(x)=6x−5,g(x)={−x+417如果x≥1,如果x<1.价值是什么 ( f ∘ g ) ( 2 ) + ( g ∘ f ) ( 0 ) (f \circ g)(2)+(g \circ f)(0) (f∘g)(2)+(g∘f)(0)? 细节和假设 ( g ∘ f ) ( x ) (g \circ f)(x) (g∘f)(x)是由 g g g而且 f f f,定义为 ( g ∘ f ) ( x ) = g ( f ( x ) ) (g \circ f)(x)= g(f(x)) (g∘f)(x)=g(f(x)).
f : R → R f: \mathbb{R} \rightarrow \mathbb{R} f:R→R函数是否满足 f ( 2 x + 1 ) = 4 x 2 + 4 x F (2x+1) = 4x^2 + 4x f(2x+1)=4x2+4x.价值是什么 f ( 16 ) f (16) f(16)?
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