PREREQUISITES:  FUNCTION REVIEW; DIFFERENCE QUOTIENTS

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Precalculus requires a thorough understanding of functions and the language used to work with functions.
The web exercises below will give you the required review.
Be sure to click-click-click through some of the exercises in each of these sections! The sections will open in a new tab/window.
Then, do the web exercise on this page to practice with difference quotients, which are particularly important in calculus.

EXAMPLE:   difference quotients

Consider a point $\,\bigl(x,f(x)\bigr)\,$ on the graph of a function $\,f\,$.
If $\,h\,$ is a small positive number, then $\,x+h\,$ lies a little to the right of $\,x\,$.
If $\,h\,$ is a small negative number, then $\,x+h\,$ lies a little to the left of $\,x\,$.
In both cases, $\,\bigl(x+h,f(x+h)\bigr)\,$ is a point on the graph of $\,f\,$, likely close to the original point $\,(x,f(x))\,$.

The slope of the line through $\,\bigl(x,f(x)\bigr)\,$ and $\,\bigl(x+h,f(x+h)\bigr)\,$ is: $$\text{slope} = \frac{\text{change in $y$}}{\text{change in $x$}} = \frac{f(x+h) - f(x)}{(x+h)-x} = \frac{f(x+h)-f(x)}{h}$$



To slide the point $\,\bigl(x+h,f(x+h)\bigr)\,$
closer to
the original point $\,\bigl(x,f(x)\bigr)\,$,

To keep the graph uncluttered,
the labels for the new (closer) points
$\bigl(x+h,f(x+h)\bigr)$
are not shown.


Recall that the result of a division problem, $\,\frac{a}{b}\,$, is called a quotient,
and the result of a subtraction problem, $\,a - b\,$, is called a difference.

The expression $\displaystyle\,\frac{f(x+h)-f(x)}{h}\,$ is called a difference quotient, because it is a quotient of differences:

In calculus, you often have to simplify difference quotients, trying to rename in a way that eliminates the $\,h\,$ in the denominator.
For example, if $\,f(x) = x^2 + 2x + 1\,$, then $$ \eqalign{ \frac{f(x+h) - f(x)}h &= \frac{\overbrace{(x+h)^2 + 2(x+h) + 1}^{f(x+h)} - (\overbrace{\vphantom{(x+h)^2} x^2 + 2x + 1}^{f(x)})}h \qquad& \text{function evaluation} \cr &= \frac{x^2 + 2xh + h^2 + 2x + 2h + 1 - x^2 - 2x - 1}{h} \qquad& \text{FOIL; distributive law}\cr &= \frac{2xh + h^2 + 2h}{h} &x^2 - x^2 = 0;\ \ 2x - 2x = 0;\ \ 1 - 1 = 0\cr &= \frac{h(2x + h + 2)}{h} = \frac hh\cdot (2x + h + 2) &\text{factor out the $h$}\cr &= 2x + h + 2 &\text{cancel: } \frac{h}{h} = 1\cr } $$ Note that the final form, $\,2x + h + 2\,$, has no $\,h\,$ in the denominator.
Note also that as $\,h\,$ gets closer and closer to $\,0\,$, the expression $\,2x + h + 2\,$ gets closer and closer to $\,2x + 2\,$.

You'll learn in calculus that $\,2x+2\,$ is called the ‘derivative of $f\,$’ and is denoted by $\,f'(x)\,$ (read as ‘$f$ prime of $\,x\,$’).
This new function $\ f'(x) = 2x + 2\ $ gives the slope of the tangent line to the graph of $\,f\,$ at the point $\,\bigl(x,f(x)\bigr)\,$!
Since $\,x\,$ can represent any real number, you might prefer to say:
This new function $\ f'(x) = 2x + 2\ $ gives the slopes of the tangent lines to the graph of $\,f\,$ at the points $\,\bigl(x,f(x)\bigr)\,$!

NOTE: The notation ‘$\,\Delta x\,$’ (read aloud as ‘delta $x$’) is often used instead of $\,h\,$ to denote a small change in $\,x\,$.
With this notation, the slope of the line through the point $\,\bigl(x,f(x)\bigr)\,$ and nearby point $\,\bigl(x+\Delta x,f(x+\Delta x)\bigr)\,$ is given by the difference quotient: $$\frac{f(x+\Delta x) - f(x)}{\Delta x}$$
When you're done practicing all the sections above
and the exercise on this page,
then move on to:
solving linear inequalities in one variable
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.