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What is the inverse of the function f (x) = 3(x + 4)^2 – 2, such that x ≤ –4?A. f^-1(x)=-4 + square root of x/3+2B. f^-1(x)=-4 - square root of x/3+2C. f^-1(x)=-4 + square root of x+2/3D. f^-1(x)=-4 - square root of x+2/3

Sagot :

Given the following function:

f(x) = 3(x + 4)^2 – 2

Let's determine its inverse form.

*Change f(x) = y, swap x and y then solve for y.

[tex]\text{ f\lparen x\rparen= 3\lparen x + 4\rparen}^2\text{ - 2}[/tex][tex]\text{ y = 3\lparen x + 4\rparen}^2\text{ - 2}[/tex][tex]\text{ x = 3\lparen y + 4\rparen}^2\text{ - 2}[/tex][tex]\text{ 3\lparen y + 4\rparen}^2\text{ - 2 = x}[/tex][tex]\text{ 3\lparen y + 4\rparen}^2\text{ = x + 2}[/tex][tex]\text{ \lparen y + 4\rparen}^2\text{ = }\frac{\text{ x + 2 }}{\text{ 3}}[/tex][tex]\text{ y + 4 = }\sqrt{\frac{\text{ x + 2 }}{\text{ 3}}}[/tex][tex]\text{y = f}^{-1}(\text{x\rparen = -4 + }\sqrt{\frac{\text{ x + 2 }}{\text{ 3 }}}[/tex]

Therefore, the answer is CHOICE C.

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