Math help plz
Find the domain and range of the following function. \[F(x) = (1+3)^{3/2}\]
Are you allowed to use a computer or calculator for this?
Neither
Yeah just use a graphing calculator
Show work
Well how are we supposed to know neither
I’m only allowed to show what i know
@TheConMan did you mean \(F(x) = (x + 3)^{3/2}\)
It specifically said in the course that if i get caught with a graphing cal, they’ll disown me
No hero. Wait, let me draw
rough course
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\(a^{c/b} = \sqrt{a^c}\)
Wait, where are we starting from? Domain? Ok
Wait, let me draw.
Well, yes, we still need to recognize what we're dealing with before we attempt to find the domain.
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Is that right?
Not exactly.
I don't know where you get the extra 3 from
That’s part of the power
You wrote two threes though.
Let me write the rule again. I don't think it was clear for you
No I didn’t. Show me.
Oh it looked like a three. It's actually a two.
.....
There's no need to put the tail on the root like that. It really did look like a three
Anyways the point is you now have a square root to deal with when finding the domain
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How do i do that?
Do you not remember what to do in this case? Set the expression under the root \( \ge 0\)
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Are you going to finish solving for \(x\)?
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So the domain is al real numbers except x has to be equal or bigger than -1
Am I right?
What da?:ö
You can describe it in words as \(x\) is all real numbers greater than or equal to -1
The domain is simply written as \([-1, \infty)\) or \(-1 \le x \le \infty\) or just \(x \ge -1\)
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You can write using set notation as well
But interval notation and inequality notation are standard.
Is my domain right tho?
Why are you trying to write it in set notation. There's many different forms of it. And it's better to just use inequality or interval notation
I jus wanna get it over with...
It depends on what your teacher prefers. I don't recognize the way you've written it. I've never seen set notation written that way. Like I said, different schools use different forms of set notation which is why I don't like it. Too many varying forms.
But inequality notation and interval notation are universally recognized
Ok let’s find the range now ::)
Just find the inverse of the function
Not inverse again...uhg ;/
Hold on, let e draw, u check
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You're supposed to square both sides.
That's how you eliminate the 2
What did i do?
Square both sides not take a root to eliminate the 2
The 2 is already a root
You perform the inverse of the root to eliminate it which is squaring
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Nope, the 1/2 is still a root
you should have \(x^2\) on the LHS
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I’m falling asleep 😴
I need a power nap
Apply the cube root to both sides
I’ll be back in a hour....dnt go. K
then subtract the 1
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What is the domain of the function?
of the inverse* that is
I can plug-in negative numbers in a cube root func right
Keep in mind we have a square root for the original function so the range will be limited by that.
I’m plug-in in numbers in the inverse func to find the dom of the original func
I mean to find range
Yes, you are but the domain of the inverse is more limited than what it appears to be
Hang on, let me see something
I say all real numbers
Well, the restriction on the range of the original function also restricts the domain of the inverse.
Yes
In order to find the lowest value of the range you have to evaluate the lower bound of the domain of the original function which is x = -1. Evaluate f(-1).
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That's the lower bound of the range
Ok what’s left
the upper bound
How
When you find the domain and range of functions you're trying to find the lower and upper bounds of both
Which is why it is best to think beyond the "all real numbers" mentality
I can’t keep my eye open...
Brb after a power nap
Take a nap. Come back later.
K let’s go.
We were looking for the upper bound. We found the lower bound.
@hero how do we find the upper bound of the graph?
So you found the inverse of the function and found that the domain would be all real numbers if the domain of the original function was not bound by the square root. And you found that the lower bound of the range of the original function is zero. This implies that the domain of the inverse is \([0, \infty)\)
So it follows that the range of the original function is ...
Give me a second to think
Trying to recall the rule, the domain of the inverse is the range of the original function... is also that [0,infinite)
If the domain of the inverse func is [0,infinite), it implies that the range of the original func is also the same
For complex problems like these, is it better to plot some damn points and find the range and domain of the function or do i have to algebraically prove everytime?
@Hero
Plotting points is always useful to get a better understanding of the behavior of the function.
It's a great go-to move if you are confused about the shape or behavior of the function. So feel free to do so.
What about tests and exams.. how should i approach the problem and btw is my range correct
Yes, the range is correct. For the exam, you want to make sure you know the general graphs of each parent function and their domain and range. You want to for example know the difference between the behavior of a cubic function and the behavior of a quadratic function.
I see k thanks vro. On to the next question.
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