Is the Inverse of f(x) a Function?

[aisha]

f(x)=5*the square root of (x+1) +2 x>=-1

determine if the inverse is a function.

I don't understand y are they telling me x>=-1? Also I know how to find the inverse but what do I do to each side of the equation to get rid of the square root? ^2? I got the inverse= [x^(2)-1]/25 I don't think this is right. How do u do this question?

 

[learningphysics]

f(x)=5*the square root of (x+1) +2 x>=-1

determine if the inverse is a function.

I don't understand y are they telling me x>=-1? Also I know how to find the inverse but what do I do to each side of the equation to get rid of the square root? ^2? I got the inverse= [x^(2)-1]/25 I don't think this is right. How do u do this question?

$$f(x)=5\sqrt{x+1}+2, x>=-1$$

The square root is only defined if the number in the root is >=0. Since x+1>=0, x>=-1.

To get rid of the square root, square both sides

To find the inverse:
$$y=5\sqrt{x+1}+2$$
$$y-2=5\sqrt{x+1}$$
$$(y-2)^2=(5\sqrt{x+1})^2$$
$$(y-2)^2=25(x+1)$$
$$x=\frac{(y-2)^2}{25}-1$$
$$f^{-1}(x)=\frac{(x-2)^2}{25}-1$$

 

[aisha]

$$f(x)=5\sqrt{x+1}+2, x>=-1$$

The square root is only defined if the number in the root is >=0. Since x+1>=0, x>=-1.

To get rid of the square root, square both sides

To find the inverse:
$$y=5\sqrt{x+1}+2$$
$$y-2=5\sqrt{x+1}$$
$$(y-2)^2=(5\sqrt{x+1})^2$$
$$(y-2)^2=25(x+1)$$
$$x=\frac{(y-2)^2}{25}-1$$
$$f^{-1}(x)=\frac{(x-2)^2}{25}-1$$

THANKS SOOOO MUCH I GET IT TOTALLY

 

[aisha]

How do u find out if the inverse is a function?

 

[quasar987]

A relation between a dependant and an independant variable is said to be a function if to each value taken by the independant variable, the dependant variable takes one and only one value.

There is a theorem that you can use that allows you to answer the question without even having to find the inverse: If a function has an interval for a domain and if it is strictly increasing, it has an inverse (i.e. the inverse is a function).

You are given the domain of your function: it is all numbers x such that $x\geq -1$. That's an interval. It's the interval $[-1, +\infty[$. So all you need to do is show that the function is strictly increasing. That is to say, we must show that for any two distinct points of the domain, say $x_1$ and $x_2$ such that $x_1 < x_2$, the two corresponding images of these two points by the function f are such that $f(x_1) < f(x_2)$.

This is not hard to do. I'll do your problem as an exemple.

Consider $x_1, x_2$ two points of the domain such that $x_1 < x_2$. This inequality implies that $f(x_1) < f(x_2)$ iff (if and only if, noted $\Leftrightarrow$)

$$5\sqrt{x_1+1}+2 < 5\sqrt{x_2+1}+2 \Leftrightarrow 5\sqrt{x_1+1} < 5\sqrt{x_2+1} \Leftrightarrow \sqrt{x_1+1} < \sqrt{x_2+1} \Leftrightarrow x_1+1 < x_2+1 \Leftrightarrow x_1 < x_2$$

,which we have supposed to be true. Therefor our proposition according to which $x_1 < x_2$ implies $f(x_1) < f(x_2)$ is true. So our function is stricly increasing and according to the theorem, the inverse is a function.


Or, if you know calculus, you can show that a function is strictly increasing by showing that the derivative is positive everywhere on the domain.

 

[learningphysics]

A particular relation is a function, if for an element in the domain, there is only one element in the range. Or in other words, for a particular x value, you only get 1 y.

For example:
y=3x+1 is a function because when you plug in an x value, you only get 1 y.

On the other hand
$$y=\pm\sqrt{x}, where\ x\geq 0$$

is not a function because, for a positive x you get 2 y values.

This is also called the vertical line test... If you plot your relation, and any vertical line you can draw intersects your plot at at most one point... then it's a function. If there's a vertical line you can draw that intersects your graph at two or more points then it's not a function.

The inverse relation in your problem is a function becuase you only get one y when you plug in an x value.

 

[aisha]

Hey guys thanks sooo much I understand now.