Question 1167052: Find the inverse function of f(x)=5x^2−15 where f^−1(110)=5. Find the domain and range of f^−1. Use interval notation.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! That's a great constraint for checking the solution! Here is the step-by-step process for finding the inverse function, its domain, and its range.
## Finding the Inverse Function, $\mathbf{f^{-1}(x)}$
The original function is $f(x) = 5x^2 - 15$. Since this is a quadratic function, it is not one-to-one on its entire natural domain ($\mathbb{R}$), meaning its inverse, $f^{-1}(x)$, would not be a function.
The condition $f^{-1}(110) = 5$ provides the necessary information to restrict the domain of $f(x)$:
1. **Interpret the Condition:** The statement $f^{-1}(110) = 5$ is equivalent to $f(5) = 110$. Let's verify this using the original function:
$$f(5) = 5(5)^2 - 15 = 5(25) - 15 = 125 - 15 = 110$$
2. **Determine the Restricted Domain of $f(x)$:** For $f(x)$ to have an inverse function, its domain must be restricted to an interval where it is one-to-one (either monotonically increasing or monotonically decreasing). Since the point $(5, 110)$ is on the graph, and the vertex of the parabola $f(x)$ is at $x=0$, we must choose the half of the parabola that contains $x=5$.
* Since $5 > 0$, we restrict the domain of $f(x)$ to **$[0, \infty)$**.
### Steps to Find $\mathbf{f^{-1}(x)}$:
1. **Replace $f(x)$ with $y$:**
$$y = 5x^2 - 15$$
2. **Swap $x$ and $y$:**
$$x = 5y^2 - 15$$
3. **Solve for $y$:**
$$x + 15 = 5y^2$$
$$\frac{x + 15}{5} = y^2$$
$$y = \pm \sqrt{\frac{x + 15}{5}}$$
4. **Select the Correct Branch:** The original function $f(x)$ was restricted to $x \ge 0$. Since the domain of $f(x)$ becomes the range of $f^{-1}(x)$, the range of $f^{-1}(x)$ must be $[0, \infty)$. Therefore, we select the positive square root.
$$f^{-1}(x) = \sqrt{\frac{x + 15}{5}}$$
***
## Domain and Range of $\mathbf{f^{-1}(x)}$
### 1. Range of $\mathbf{f^{-1}(x)}$
The range of the inverse function is the restricted domain of the original function.
$$\text{Range}(f^{-1}) = \text{Domain}(f) = [0, \infty)$$
### 2. Domain of $\mathbf{f^{-1}(x)}$
The domain of the inverse function is the range of the original function.
* The restricted domain of $f(x)$ is $[0, \infty)$.
* The function is $f(x) = 5x^2 - 15$.
* The minimum value (the vertex) occurs at $x=0$:
$$f(0) = 5(0)^2 - 15 = -15$$
* As $x \to \infty$, $f(x) \to \infty$.
* Thus, the range of $f(x)$ is $[-15, \infty)$.
* Therefore, the domain of $f^{-1}(x)$ is $[-15, \infty)$. (This also ensures the expression under the square root, $\frac{x+15}{5}$, is non-negative.)
***
## Summary
The inverse function is:
$$f^{-1}(x) = \sqrt{\frac{x + 15}{5}}$$
| Property | Interval Notation |
| :--- | :--- |
| **Domain of $\mathbf{f^{-1}}$** | **$[-15, \infty)$** |
| **Range of $\mathbf{f^{-1}}$** | **$[0, \infty)$** |
Answer by ikleyn(52915)  (Show Source):
You can put this solution on YOUR website! .
Find the inverse function of f(x)=5x^2−15 where f^−1(110)=5.
Find the domain and range of f^−1. Use interval notation.
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Hello, @CPhill,
I am glad to see your posts at this forum again.
It tells me that you accept my criticism and find it useful.
It is good and promises a fruitful cooperation.
Now I would like to discuss your post and your solution to this problem.
Formally speaking, the solution by @CPhill is correct, but it is difficult to read
and difficult for a student to learn from it.
It is because (a) the tone is too formal, and (b) in many places, the same thought
is repeated more than one time (~ several times).
Actually, repeating is a good pedagogical technique for conveying a thought or idea to a student's mind.
But then a teacher or a professor should clearly emphasize (~ underline) that they are repeating - then it will be better
received by the audience.
Therefore, I will present here a way, as I would explain a solution.
In this problem, f(x) = 5x^2 - 15 represents a parabola opened upward with the vertex point (0,-15)
and the symmetry axis x= 0.
From this description, it is clear that f(x) has two branches, where the function is monotonic.
One branch is in the domain 0 <= x < infinity, and the other branch is in the domain -infinity < x <= 0.
Function f^-1(x) will be inverse to one of these two branches, and we should decide - to which one.
To mdetermine it, look at the condition f^-1(110) = 5.
It is the same as to say f(5) = 110, and it tells us that this function f^-1 is inverse to the right
branch of function f(x), because "a 5" belongs to this branch.
OK, very good. Now we want to express the inverse function f^-1(x) explicitly.
It can be done in two ways, on the student' choice.
One way is formal.
We write y = 5x^2 - 15 and swap x and y in this formula
x = 5y^2 - 15.
Then we express 'y' from this formula
x + 15 = 5y^2,
y^2 =  ,
y = +/-  . (1)
We are almost done - we only need to determine, which sign to use at the square root, PLUS or MINUS.
To determine it, recall the given condition f^-1(110) = 5.
It tells us to substitute x= 110 into formula (1).
You will get then y = +/-  = +/-  = +/-  = +/- 5.
But the condition says that f^-1(110) = 5 - so, it dictates us to choose the sign ' + '.
Thus finally the formula for f^-1(x) is
f^-1(x) = . (2)
You may check that f^-1(110) = 5 by substituting x=110 into this formula (as i did it in couple of lines above).
So, it is how the formal way proceeds.
Another way is informal.
We see that the direct function squares x, then multiplies the square by 5 and subtracts 15.
The inverse function will make inverse operations in reverse order.
It will add 15 to the argument, will divide the sum by 5 and will take square root - - - exactly as our formula (2) does.
At this point, my solution is complete.
Enjoy new knowledge you learned from this my lecture.
---------------------------------------
The tone is important feature of teaching on problem solving.
Remember that you are before your audience to teach - not to demonstrate your superiority.
The tone should not be as in an encyclopedia or in a handbook.
The tone of writing here should be as at normal communication/conversation - as if you re-tell a story.
Sometimes, it is good to insert couple of words, like "now we completed the first part
and are switching to the second one", to broke a monotonic narration in parts,
that are easier to absorb.
What I am saying does not mean to give in severity, but means to be like a human -
- not like a computer retelling encyclopedia.
. . . . . . . . . . . . . . . .
Our job as teachers is not only to teach problem solving, but also to leave students
with a positive impression of the subject, and perhaps even to engage them.
This is just the ultimate goal.
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