Find The Range And Inverse (please Help I Have Test Monday)

AI Thread Summary
To find the range and inverse of the function f(x) = √[(8x² + 1)/(9 - 5x²)], it's important to understand that the range consists of all possible values of f(x). The inverse function can be determined by swapping x and y in the equation, although solving for y may not always be feasible. The range can be calculated by ensuring the radicand remains non-negative, which establishes the restrictions for y. The function primarily involves x² terms, simplifying the process of finding the range. Understanding these concepts is crucial for test preparation.
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I don't understand this at all...

Find the Range and Inverse of f(x) = square root of [(8x^2 +1)/ (9-5x^2)]


I am so confused please help!
 
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The range of a function is all the values f(x) can take. The inverse of a function is basically the function that cancels it out. The inverse of division by 3 is multiplication by 3. The inverse of square rooting a positive number is squaring that number. To find the inverse function when you have a general y=f(x) is to swap the y's and x's, ie x=f(y), and it is not necessarily required to solve again for y, and in general it is not even possible in terms of elementary functions, but sometimes you lose marks when you can but you don't.
 
The range is easily solvable because only x^2 terms exist. Solving for x in terms of y can thus be done by a simple square root. Just make sure the radicand is not negative and that will give you the restrictions for the range y.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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