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Find the function f (x) = {X & # 178; 2x-3 (- 2 ≤ x)
f(x)={x² -2x-3 (-2≤x
Given f (2 √ x + 1) = x & # 178; - 2x, then f (x) =?
It has a root
F (2 radical x + 1) = x & # 178; - 2x
Given f [2 √ (x + 1)] = x & # 178; - 2x, then f (x) =? (is the title like this?)
Let 2 √ (x + 1) = u, then x = (U & # 178 / 4) - 1, and substitute it into the original formula to get:
f(u)=[(u²/4)-1]²-2[(u²/4)-1]=(u⁴/16)-(u²/2)+1-(u²/2)+2=(u⁴/16)-u²+3
If u is replaced by X, f (x) = (1 / 16) x & # 8308; - X & # 178; + 3
Given f (2x-1) = x & # 178;, then f (x) =?
2, given the function FX = KX + 5, if f (2) = 3, then the value range of X with FX > 0 is --
Given f (2x-1) = x & # 178;, let 2x-1 = t; X = (T + 1) / 2; then f (x) = (x + 1) &# 178 / / 4; 2, given function FX = KX + 5, if f (2) = 3, then the value range of X with FX > 0 is x ＜ 5.2k + 5 = 3; k = - 1; ■ - x + 5 ＞ 0; X ＜ 5; I'm glad to answer for you, skyhunter 002 will answer your questions
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