Find and simplify the difference quotient for f(x) = x-7
(I honestly have no idea what that question is asking for?)
2. Find (h o g)(2) for g(x)=-3x , h(x)= x^3-2
I'm also not sure about the first question, but for the second one:
$\displaystyle h\circ g(x)=h(g(x))$
So you have:
$\displaystyle h(-3x)$
$\displaystyle =(-3x)^3-2$
Now plug in 2 in place of $\displaystyle x$ :
$\displaystyle (-3\times2)^3-2$
$\displaystyle =-218$
The difference quotient of the function $\displaystyle f(x)$ is given by $\displaystyle \frac{f(x + \Delta x) - f(x)}{\Delta x}$.
If $\displaystyle f(x) = x - 7$, then $\displaystyle f(x + \Delta x) = x + \Delta x - 7$.
Substitute $\displaystyle f(x + \Delta x) = x + \Delta x - 7$ and $\displaystyle f(x) = x - 7$ into the difference quotient.
$\displaystyle \frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{x + \Delta x - 7 - (x - 7)}{\Delta x} = \frac{x + \Delta x - 7 - x + 7}{\Delta x} = \frac{\Delta x}{\Delta x} = 1$
Not precisely. Certainly to do that you would need to find the "difference quotient" but it is possible to find the difference quotient ("average rate of change") without taking the limit or finding the derivative. Often such a problem is given as a prelude to introducing the derivative.