Question

Find the Inverse of these Functions:

f(x)= square root x + 1

f(X) =x^2 + x - 1

f(x) = 2x - 4

f(x)= (4-x) / 3+x

Answer 1

Answer on Question #44020 – Engineering – Other

Find the Inverse of these Functions:

$$f(x) = \text{square root } x + 1$$ $$f(X) = x^2 + x - 1$$ $$f(x) = 2x - 4$$ $$f(x) = (4 - x) / 3 + x$$

Solution:

Given the function $f(x)$ we want to find the inverse function, $f^{-1}(x)$.

1. First, replace $f(x)$ with $y$. This is done to make the rest of the process easier.

2. Replace every $x$ with a $y$ and replace every $y$ with an $x$.

3. Solve the equation from Step 2 for $y$. This is the step where mistakes are most often made so be careful with this step.

4. Replace $y$ with $f^{-1}(x)$. In other words, we’ve managed to find the inverse at this point.

#1

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

Next, replace all $x$'s with $y$ and all $y$'s with $x$:

$$x = \sqrt{y} + 1$$

Now, solve for $y$:

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

Finally replace $y$ with $f^{-1}(x)$.

$$f^{-1}(x) = (x - 1)^2, x \geq 0$$

#2

$$f(x) = x^2 + x - 1$$ $$y = x^2 + x - 1$$

Next, replace all $x$'s with $y$ and all $y$'s with $x$:

$$x = y^2 + y - 1$$

Now, solve for $y$:

$$y^2 + y - (1 + x) = 0$$ $$y = \frac{-1 \pm \sqrt{1 + 4(1 + x)}}{2}$$

Finally replace $y$ with $f^{-1}(x)$.

$$f^{-1}(x) = \frac{-1 \pm \sqrt{5 + 4x}}{2}, x \geq -\frac{5}{4}$$

#3

$$\begin{array}{l} f(x) = 2x - 4 \\ y = 2x - 4 \\ \end{array}$$

Next, replace all x's with y and all y's with x:

$$x = 2y - 4$$

Now, solve for y:

$$\begin{array}{l} 2y = x + 4 \\ y = \frac{x + 4}{2} \\ \end{array}$$

Finally replace y with $f^{-1}(x)$.

$$f^{-1}(x) = \frac{x + 4}{2}$$

4

$$f(x) = \frac{4 - x}{3} + x$$ $$y = \frac{4 - x}{3} + x$$

Next, replace all x's with y and all y's with x:

$$x = \frac{4 - y}{3} + y$$

Now, solve for y:

$$\begin{array}{l} 3x = 4 - y + 3y \\ 3x = 4 + 2y \\ y = \frac{3x - 4}{2} \\ \end{array}$$

Finally replace y with $f^{-1}(x)$.

$$f^{-1}(x) = \frac{3x - 4}{2}$$

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