What is the inverse of f(x)=x^2+3,x<= 0 ?

The inverse of f(x)=x^2+3,x<= 0 is sqrt(x-3),-sqrt(x-3)

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inverse of f(x)=x^2+3,x<= 0

inverse

f (x) = x^{2} + 3, x \leq 0

x - 3, - x - 3

Solution steps

y = x^{2} + 3

Replace

x

with

y

x = y^{2} + 3

Solve for

y, x = y^{2} + 3

x - 3, - x - 3

f (x) = \frac{x ^{2} + 6 x - 16}{x ^{2} - 4}

x^{2} - 4 x + 8

f (x) = \frac{7 x ^{2} + 16}{x ^{2} - 16}

(8.1) (- 4.2)

4 sin (x)

What is the inverse of f(x)=x^2+3,x<= 0 ?
The inverse of f(x)=x^2+3,x<= 0 is sqrt(x-3),-sqrt(x-3)