How do you find the inverse of $f (x) = - \sqrt{2 x + 1}$ $f(x)= -sqrt(2x+1)$ ?

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Answer 1 · 2018-03-31T13:55:14

$f^{- 1} (x) = \frac{1}{2} x^{2} - \frac{1}{2}$ $f^-1(x)=1/2x^2-1/2$

To find the inverse we need to express $x$ $x$ as a function of $y$ $y$ :

$y = - \sqrt{2 x + 1}$ $y=-sqrt(2x+1)$

Square both sides:

$y^{2} = 2 x + 1$ $y^2=2x+1$

$y^{2} - 1 = 2 x$ $y^2-1=2x$

$x = \frac{y^{2} - 1}{2}$ $x=(y^2-1)/2$

$x = \frac{1}{2} y^{2} - \frac{1}{2}$ $x=1/2y^2-1/2$

Substitute:

$y = x$ $y=x$

$y = \frac{1}{2} x^{2} - \frac{1}{2}$ $y=1/2x^2-1/2$

$:.$

$f^-1(x)=1/2x^2-1/2$

Notice that the negative half of the graph of $1/2x^2-1/2$ is the graph of

$-sqrt(2x+1)$ reflected in the line $y=x$ . This is typical for the inverse of a function.

GRAPH:

enter image source here

How do you find the inverse of f(x)= -sqrt(2x+1)f(x)=−√2x+1f(x)= -sqrt(2x+1)?