The discussion focuses on simplifying complex equations involving exponents and square roots. A user expresses confusion about squaring the expression "square root of x over 2x" and seeks clarification on how to apply functions f(x) and g(x). Another participant suggests substituting g(x) into f(x) to find f(g(x)), addressing the user's misunderstanding. The user also mentions frustration with a lack of correct multiple-choice answers and difficulty adjusting to non-calculator math. Overall, the thread emphasizes the challenges of understanding exponent rules and function composition without calculator assistance.
#1
dustinj.11
2
0
Homework Statement
Attached is the problem, I don't like the math editor used on this site, so its a pdf. The only real problem I have is how to is how do you square a problem like square root of x over 2x?
Attached is the problem, I don't like the math editor used on this site, so its a pdf. The only real problem I have is how to is how do you square a problem like square root of x over 2x?
Welcome to the PF.
Since you are given f(x) = \frac{1}{\sqrt{x}}
and g(x) = x^2 - 5
what is f(g(x))? Wouldn't you substitute the g(x) term into the f(x) equation? Where do you get "square root of x over 2x"?
#3
dustinj.11
2
0
Thanks for the help, bt the answers were multiple choice, and as it turns out the correct choice wasn't there. Thats what confused me. Sorry if i seemed a little confused, just I was bred on a calculator, and now my calc ab teacher won't let us use any type of calculator. So, 90% of the time my brain is fried.
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