Inverse ahh

[aisha]

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

determine if the inverse is a function.

I dont 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 dont think this is right. How do u do this question??? :confused: :cry:

[learningphysics]

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

determine if the inverse is a function.

I dont 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 dont think this is right. How do u do this question??? :confused: :cry:

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. :smile: