Need help finding the range of a function.

[saladtim]

Looking for someone to point me in the right direction here. I need to find the range of the function. I tried taking the inverse but quickly realized that was not going to work. Again, i'm just looking for some general direction here, where to start or which method to use, not the whole answer.

 

[MarkFL]

Can you identify the domain of the given expression?

 

[Steven G]

Set the equation equal to K and then solve for x.

Now this is the x that when you plug into this equation will give you k, so k is in the range. Now for example if you solve for x and get x= 1/(k-3). Now there is no x when k=3. So k is NOT in the domain.

 

[Dr.Peterson]

I think solving for x will require solving a quartic equation, so that probably won't work well.

I would try sketching the graph, using the derivative and other tools to find local maxima or minima, which can be put together to find the range.

 

[MarkFL]

I think solving for x will require solving a quartic equation, so that probably won't work well.

I would try sketching the graph, using the derivative and other tools to find local maxima or minima, which can be put together to find the range.

Yes, that's there I was eventually going to head, after using the domain the identify vertical asymptotes, etc.

 

[saladtim]

Thanks for the responses everyone!

Can you identify the domain of the given expression?

This is what i came up with for the domain. I think I was trying too hard to find an algebraic solution, I will have to try a graphical approach now.
Thanks again.

 

[Otis]

… This is what i came up with for the domain …

Hello. Good job, but you've got a bit more work to do, on that domain. So far, your result shows all values that won't lead to zero in the function's denominator. Next, you need to eliminate values that make the numerator undefined. For example, what happens if you try to evaluate the function for x=-4?

Are you allowed to use calculus, for finding the range? If not, I'm wondering whether you're allowed to use technology (like a graphing calculator) or a numerical method, to approximate a solution for the range.
?

Edit: Grammar typo

 

[saladtim]

Next, you need to eliminate values that make the numerator undefined. For example, what happens if you try to evaluate the function for x=-4?

right thank you I totally skipped that step. going to reevaluate the domain.

Are you allowed to use calculus, for finding the range? If not, I'm wondering whether you're allowed to technology (like a graphing calculator) or a numerical method, to approximate a solution for the range.
?

We are not using calculus yet, but graphing calculators are allowed.

 

[MarkFL]

...We are not using calculus yet, but graphing calculators are allowed.

Hmm...without the calculus it will be more difficult to analytically find the range. But, if graphing software/calculators are allowed you can at least approximate it very well.

 

[saladtim]

alright that should be solid.

 

[saladtim]

Hmm...without the calculus it will be more difficult to analytically find the range...

Maybe I am wrong about calculus not being allowed. I don't think the answer is wanted as a decimal as we are putting it into interval notation.
I don't think we have learned how to solve with calculus, I think I am expected to get this information from the graph.

 

[Otis]

… graphing calculators are allowed.

Oh, good. You can zoom in on that local maximum point (between x-values -2/3 and 1/2).
?

 

[saladtim]

how does this look?

 

[Otis]

Looks good to me, but I would replace the 'equal to' sign in the range statement to \(\displaystyle \approx\) because the decimal number is approximated.
?

 

[MarkFL]

W|A gives a value of:

[MATH]y=\frac{9\sqrt{23+\sqrt{1969}}}{911-23\sqrt{1969}}\approx-0.674095748943020[/MATH]
for that local maximum in the middle.

 

[saladtim]

Looks good to me, but I would replace the 'equal to' sign in the range statement to \(\displaystyle \approx\) because the decimal number is approximated.
?

Right that would be more accurate, thanks.

W|A gives an value of:

[MATH]y=\frac{9\sqrt{23+\sqrt{1969}}}{911-23\sqrt{1969}}\approx-0.674095748943020[/MATH]
for that local maximum in the middle.

Okay yeah that is definitely not something we have learned, but I honestly appreciate the time you've put into helping me out here. Both you and Otis have helped me get a much better understanding of this problem. You guys are seriously rock stars.

 

[mmm4444bot]

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