Simplifying Difference Quotients with Variable Expressions - How to Solve?

AI Thread Summary
To solve the difference quotient for the function f(x) = (5x)/(x-1), first substitute a into the function to find f(a) = (5a)/(a-1) and then substitute a + h to find f(a + h) = (5(a + h))/((a + h) - 1). The difference quotient is calculated as (f(a + h) - f(a))/h. It's important to use parentheses correctly during substitution to avoid confusion. Properly substituting the expressions will clarify the calculations needed to solve the problem.
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Homework Statement


Find f(a), f(a + h),and the difference quotient f(a + h) − f(a), h where h ≠ 0.

f(x) = 5x/x-1

f(a) = ?
f(a + h) = ?
f(a + h) − f(a)/(h) = ?


The Attempt at a Solution



As I understand it I must substitute the values but I don't understand the question. say if it were something like 1/x I would know to fill it into (1/(x+h)-1/x) / h
But this just confuses me I don't know how to answer these questions
 
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Niaboc67 said:

Homework Statement


Find f(a), f(a + h),and the difference quotient f(a + h) − f(a), h where h ≠ 0.

f(x) = 5x/x-1

f(a) = ?
f(a + h) = ?
f(a + h) − f(a)/(h) = ?


The Attempt at a Solution



As I understand it I must substitute the values but I don't understand the question. say if it were something like 1/x I would know to fill it into (1/(x+h)-1/x) / h
But this just confuses me I don't know how to answer these questions

The way you wrote it you have
f(x) = \frac{5x}{x} - 1 = 5 - 1 = 4
so ##f(a) = f(a+h) = 4##. Perhaps you really did not mean what you wrote, but because you failed to use parentheses it is impossible to say. And, of course, you wrote
f(a+h) - \frac{f(a)}{h} = 4 - \frac{4}{h}.
Again, maybe you meant something different.
 
What I meant was: f(x) = (5x)/(x-1) my mistake
 
Niaboc67 said:
What I meant was: f(x) = (5x)/(x-1) my mistake
Just replace every instance of x with the given value. Where the given value consists of an expression, like a+h, put the expression in parentheses when making the substitution.
 
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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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