4.08 Inverse functions

To find the inverse of the function represented by the table of values, we need to swap the x and y-values. Then, create a new table of values representing the inverse function, f^{-1}\left(x\right).

The table of values for the inverse function is:

To verify the functions represented by the table of values are inverses, we can plot the points on a coordinate plane and see if the points are mirror images over the line y=x.

f(x)=\sqrt[3]{x}

To find the inverse of the function graphically, we need to swap the x and y-coordinates and reflect the original function across the line y=x. We first need to identify points that lie on the function.

Reflecting the original function across the line y=x by swapping the x and y-coordinates, we get the following inverse function:

The inverse function is a polynomial function that we have seen before. The inverse of the original function, f\left(x\right)=\sqrt[3]{x}, is f^{-1}(x) = x^3.

f(x)=\sqrt{x}+3

To find the inverse of the function graphically, we need to identify points on the original function and swap the x and y-coordinates to reflect it across the line y=x.

Reflecting the original function across the line y=x, we get the following inverse function:

We have graphed the inverse function, but let's now take a look at the domain and range of our function in comparision to the inverse.

For the function f(x), we have a domain of [0, \infty) and a range of [3, \infty).

The inverse function, f^{-1}(x), has a domain of [3, \infty) and a range of [0, \infty).

We can clearly see that the domain of our function has become the range of our inverse and the range of our function has become the domain of the inverse.

We can also identify the inverse function equation, which is f^{-1}(x) = (x - 3)^2 for x\geq 3.

f\left(x\right)

To determine if a function is invertible, we will reflect the graph over the line y=x, then use the vertical line test to determine if the inverse is a function.

In addition, we can see that the inverse is a line, which indicates it is a linear function.

As the two functions are reflections of each other across the line y=x, we could apply the horizontal line test on f\left(x\right).

If we draw a horizontal line anywhere on the graph and the horizontal line only intersects the function once, then the inverse of the function will also be a function, without any domain restrictions.

g\left(x\right)

To determine if a function is invertible, we will reflect the graph over the line y=x, then use the vertical line test to determine if the inverse is a function.

Therefore, g\left(x\right) does not have an inverse function.

Alternatively, we could perform the horizontal line test on g\left(x\right), revealing the function does not have an inverse.

However, if g\left(x\right) had a restricted domain of either \left[0, \infty\right) or \left( -\infty, 0\right], then it would have an inverse function, as shown for \left[0, \infty\right).

h\left(x\right)

To determine if a function is invertible, we will reflect the graph over the line y=x, then use the vertical line test to determine if inverse is a function.

It appears that the inverse is a cube root function, which means there is only one y-value when x=-3. This confirms that the original function h\left(x\right) is invertible.

y=7x- 4

To find the inverse, we will swap x and y, then solve for y. We can determine whether the inverse is a function by using the vertical line test.

y=\dfrac{x}{7}+\dfrac{4}{7} is the inverse of y=7x-4.

We now need to determine if y=\dfrac{x}{7}+\dfrac{4}{7} is a function. Graphing y=\dfrac{x}{7}+\dfrac{4}{7} using technology, the relation passes the vertical line test and therefore is a function.

y = \left(x-3\right)^2-5

We want to swap x and y, then solve for y. We can then check whether the inverse is a function using the vertical line test.

Inverse:

y=3\pm\sqrt{x+5} is the inverse of y=\left(x-3\right)^2-5.

We now need to determine if y=3\pm\sqrt{x+5} is a function. Graphing y=3\pm\sqrt{x+5} using technology, it is clear to see that it does not pass the vertical line test and therefore is not a function.

We can restrict the domain of y= \left(x - 3 \right) ^2 - 5 to x \geq 3, and it will have the inverse function y= 3 + \sqrt{x+5}.

As the graph of a quadratic function does not pass the horizontal line test, no quadratic functions have inverses without domain restrictions.

Alternatively, we could have restricted the domain to x\leq 3, in which case the inverse would have been y=3-\sqrt{x+5}.

f \left(x \right) = \sqrt{x-3} +2 for x\geq3

Since this is written as a function, we will begin by writing f\left(x\right) as y. Then, we can swap x and y, solve for y, and determine whether the inverse is a function by using the vertical line test.

We begin by writing f\left(x\right) as y. So, we have y=\sqrt{x-3}+2.

Next, we swap x and y, resulting in x=\sqrt{y-3}+2.

Now, we need to solve for y.

We must now take a look at domain and range. We know with inverse functions, our domain and range are swapped.

• The domain of f(x) was [3, \infty), which means that this interval is also the range of f^{-1}(x).

• The range of f(x) was [2,\infty), which means that this interval is also the domain of the inverse, f^{-1}(x).

The inverse function is f^{-1}(x)=(x-2)^2+3 with a domain of [2,\infty) and a range of [3,\infty).

We now need to determine if y=(x-2)^2+3 is a function. Graphing y=(x-2)^2+3 over x\geq 3 using technology, it is clear to see that it passes the vertical line test and therefore is a function.

f(x)=4x-2 and g(x)=\dfrac{x+2}{4}

To determine if the given functions are inverses of each other, we can use the idea that two functions, f(x) and g(x), are inverses if f(g(x)) = g(f(x)) = x. We will find both f(g(x)) and g(f(x)) and check if they are equal to x.

First, let's find f(g(x)).

Now, let's find g(f(x)).

Since f(g(x)) = x and g(f(x)) = x, we can conclude that the given functions are inverses of each other.

To verify our results graphically, we can use graphing software or a graphing calculator to plot the functions f(x) and g(x) on the same set of axes. If the functions are inverses, their graphs should be reflections of each other across the line y=x.

When graphing the functions, we can observe that they are indeed reflections of each other across the line y=x, confirming that they are inverses.

f(x)=5(x+4)^3 and g(x)=\dfrac{1}{5}\sqrt[3]{x}-4

To determine if the given functions are inverses of each other, we will use the idea that two functions, f(x) and g(x), are inverses if f(g(x)) = g(f(x)) = x. We will compute f(g(x)) and g(f(x)) to check if they both simplify to x.

First, let's compute f(g(x)):

Since f\left(g\left(x\right)\right)\neq x, we do not need to compute g(f(x)). The functions are not inverses of each other.

To verify our result graphically, we can use graphing software or a graphing calculator to plot the functions f(x) and g(x) on the same set of axes. Upon graphing the functions, we can see that they are not reflections of each other across the line y=x, indicating that they are not inverses.

f(x)=\sqrt{x-5} and g(x)=x^2 +5, x \geq 5

To determine if the given functions are inverses, we need to verify if f(g(x)) = g(f(x)) = x. By substituting the expression of g(x) into f(x) and vice versa, we can check if the resulting expressions simplify to x.

First, let's find f(g(x)):

Now, let's find g(f(x)):

Since f(g(x)) = g(f(x)) = x, the given functions are inverses of each other.

Let's algebraically compute the inverse of each function and compare the results to verify if they are inverses of each other.

First, let's find the inverse of f(x)=\sqrt{x-5}.

The inverse of f(x) is f^{-1}(x)=x^2 + 5.

Now, let's find the inverse of g(x)=x^2 +5.

The inverse of g(x) is g^{-1}(x)=\sqrt{x-5}.

Comparing the original functions with their inverses, we can see that f(x) is the inverse of g(x) and vice versa. Therefore, the given functions are inverses of each other.