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Solved Examples on Difference Quotient

You may verify your answers as applicable with the: Relations and Functions Calculators
Calculate the difference quotient for the functions
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$DQ = \dfrac{f(x + h) - f(x)}{h}$

(1.) $f(x) = x$

$f(x) = x \\[3ex] f(x + h) = x + h \\[3ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = \dfrac{x + h - x}{h} \\[5ex] DQ = \dfrac{h}{h} \\[5ex] DQ = 1$
(2.) $f(x) = -x$

$f(x) = -x \\[3ex] f(x + h) = -(x + h) \\[3ex] f(x + h) = -x - h \\[3ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = \dfrac{-x - h - (-x)}{h} \\[5ex] DQ = \dfrac{-x - h + x}{h} \\[5ex] DQ = \dfrac{-h}{h} \\[5ex] DQ = -1$
(3.) $f(x) = -x^2$

$f(x) = -x^2 \\[3ex] f(x + h) = -(x + h)^2 \\[3ex] f(x + h) = -(x + h)(x + h) \\[3ex] f(x + h) = -(x^2 + hx + hx + h^2) \\[3ex] f(x + h) = -(x^2 + 2hx + h^2) \\[3ex] f(x + h) = -x^2 - 2hx - h^2 \\[3ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = \dfrac{-x^2 - 2hx - h^2 - (-x^2)}{h} \\[5ex] DQ = \dfrac{-x^2 - 2hx - h^2 + x^2)}{h} \\[5ex] DQ = \dfrac{-2hx - h^2}{h} \\[5ex] DQ = \dfrac{h(-2x - h)}{h} \\[5ex] DQ = -2x - h$
(4.) $f(x) = x^2$

$f(x) = x^2 \\[3ex] f(x + h) = (x + h)^2 \\[3ex] f(x + h) = (x + h)(x + h) \\[3ex] f(x + h) = x^2 + hx + hx + h^2 \\[3ex] f(x + h) = x^2 + 2hx + h^2 \\[3ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = \dfrac{x^2 + 2hx + h^2 - x^2}{h} \\[5ex] DQ = \dfrac{2hx + h^2}{h} \\[5ex] DQ = \dfrac{h(2x + h)}{h} \\[5ex] DQ = 2x + h$
(5.) $f(x) = x^3$

$f(x) = x^3 \\[3ex] f(x + h) = (x + h)^3 \\[3ex] f(x + h) = x^3 + 3x^2h + 3xh^2 + h^3 \\[3ex]$
Student: That was fast.
How did you get it?
Teacher: I used the Pascal's Triangle.
Student: What is it?
Teacher: Here is the video: Pascal's Triangle and the Binomial Theorem
You can expand binomials raised to any positive exponent quickly using that method.
Would you like for me to do it another way?

$(x + h)^3 = (x + h)(x + h)(x + h) \\[3ex] (x + h)(x + h) = x^2 + hx + hx + h^2 \\[3ex] (x + h)(x + h) = x^2 + 2hx + h^2 \\[3ex] (x^2 + 2hx + h^2)(x + h) = x^3 + hx^2 + 2hx^2 + 2h^2x + h^2x + h^3 \\[3ex] (x + h)^3 = x^3 + 3x^2h + 3hx^2 + 3h^2x + h^3 \\[3ex] (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \\[3ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = \dfrac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} \\[5ex] DQ = \dfrac{3x^2h + 3xh^2 + h^3}{h} \\[5ex] DQ = \dfrac{h(3x^2 + 3xh + h^2)}{h} \\[5ex] DQ = 3x^2 + 3xh + h^2$
(6.) $f(x) = 3x - 7$

$f(x) = 3x - 7 \\[3ex] f(x + h) = 3(x + h) - 7 \\[3ex] f(x + h) = 3x + 3h - 7 \\[3ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = \dfrac{3x + 3h - 7 - (3x - 7)}{h} \\[5ex] DQ = \dfrac{3x + 3h - 7 - 3x + 7}{h} \\[5ex] DQ = \dfrac{3h}{h} \\[5ex] DQ = 3$
(7.) $f(x) = -x^3$

$f(x) = -x^3 \\[3ex] f(x + h) = -(x + h)^3 \\[3ex] f(x + h) = -1(x^3 + 3x^2h + 3xh^2 + h^3) \\[3ex] f(x + h) = -x^3 - 3x^2h - 3xh^2 - h^3 \\[3ex]$
We can also do multiply the two binomials first; then multiply the resulting trinomial by the binomial.
Then, we multiply the product by $-1$
$-(x + h)^3 = -1 * (x + h)(x + h)(x + h) \\[3ex] (x + h)(x + h) = x^2 + hx + hx + h^2 \\[3ex] (x + h)(x + h) = x^2 + 2hx + h^2 \\[3ex] (x^2 + 2hx + h^2)(x + h) = x^3 + hx^2 + 2hx^2 + 2h^2x + h^2x + h^3 \\[3ex] (x + h)^3 = x^3 + 3x^2h + 3hx^2 + 3h^2x + h^3 \\[3ex] (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \\[3ex] -(x + h)^3 = -1 * (x^3 + 3x^2h + 3xh^2 + h^3) \\[3ex] -(x + h)^3 = -x^3 - 3x^2h - 3xh^2 - h^3 \\[3ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = \dfrac{-x^3 - 3x^2h - 3xh^2 - h^3 - (-x^3)}{h} \\[5ex] DQ = \dfrac{-x^3 - 3x^2h - 3xh^2 - h^3 + x^3)}{h} \\[5ex] DQ = \dfrac{-3x^2h - 3xh^2 - h^3}{h} \\[5ex] DQ = \dfrac{h(-3x^2 - 3xh - h^2)}{h} \\[5ex] DQ = -3x^2 - 3xh - h^2$
(8.) $f(x) = -3x + 12$

$f(x) = -3x + 12 \\[3ex] f(x + h) = -3(x + h) + 12 \\[3ex] f(x + h) = -3x - 3h + 12 \\[3ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = \dfrac{-3x - 3h + 12 - (-3x + 12)}{h} \\[5ex] DQ = \dfrac{-3x - 3h + 12 + 3x - 12}{h} \\[5ex] DQ = \dfrac{-3h}{h} \\[5ex] DQ = -3$
(9.) $f(x) = -10x^2 + 7x + 5$

$f(x) = -10x^2 + 7x + 5 \\[3ex] f(x + h) = -10(x + h)^2 + 7(x + h) + 5 \\[3ex] f(x + h) = -10[(x + h)(x + h)] + 7x + 7h + 5 \\[3ex] f(x + h) = -10(x^2 + hx + hx + h^2) + 7x + 7h + 5 \\[3ex] f(x + h) = -10(x^2 + 2hx + h^2) + 7x + 7h + 5 \\[3ex] f(x + h) = -10x^2 - 20hx - 10h^2 + 7x + 7h + 5 \\[3ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = \dfrac{-10x^2 - 20hx - 10h^2 + 7x + 7h + 5 - (-10x^2 + 7x + 5)}{h} \\[5ex] DQ = \dfrac{-10x^2 - 20hx - 10h^2 + 7x + 7h + 5 + 10x^2 - 7x - 5}{h} \\[5ex] DQ = \dfrac{-20hx - 10h^2 + 7h}{h} \\[5ex] DQ = \dfrac{h(-20x - 10h + 7)}{h} \\[5ex] DQ = -20x - 10h + 7$
(10.) $f(x) = -\dfrac{1}{16x}$

$f(x) = -\dfrac{1}{16x} \\[5ex] f(x + h) = -\dfrac{1}{16(x + h)} \\[5ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex]$ Let us do this one at a time - piecemeal approach
Let us deal with the numerator first.

$Numerator = f(x + h) - f(x) \\[3ex] f(x + h) - f(x) = -\dfrac{1}{16(x + h)} - -\dfrac{1}{16x} \\[5ex] = -\dfrac{1}{16(x + h)} + \dfrac{1}{16x} \\[5ex] = -\dfrac{x}{16x(x + h)} + \dfrac{1(x + h)}{16x(x + h)} \\[5ex] = -\dfrac{x}{16x(x + h)} + \dfrac{x + h}{16x(x + h)} \\[5ex] = \dfrac{-x + x + h}{16x(x + h)} \\[5ex] = \dfrac{h}{16x(x + h)} \\[5ex] Denominator = h \\[3ex] DQ = \dfrac{Numerator}{Denominator} \\[5ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = [f(x + h) - f(x)] \div h \\[5ex] DQ = \dfrac{h}{16x(x + h)} \div \dfrac{h}{1} \\[5ex] DQ = \dfrac{h}{16x(x + h)} * \dfrac{1}{h} \\[5ex] DQ = \dfrac{1}{16x(x + h)}$
(11.) $f(x) = \dfrac{x}{x + 1}$

$f(x) = \dfrac{x}{x + 1} \\[5ex] f(x + h) = \dfrac{x + h}{x + h + 1} \\[5ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex]$ Let us do this one at a time - piecemeal approach
Let us deal with the numerator first.

$Numerator = f(x + h) - f(x) \\[3ex] f(x + h) - f(x) = \dfrac{x + h}{x + h + 1} - \dfrac{x}{x + 1} \\[5ex] = \dfrac{(x + h)(x + 1)}{(x + h + 1)(x + 1)} - \dfrac{x(x + h + 1)}{(x + h + 1)(x + 1)} \\[5ex] (x + h)(x + 1) = x^2 + x + hx + h \\[3ex] x(x + h + 1) = x^2 + hx + x \\[3ex] = \dfrac{x^2 + x + hx + h}{(x + h + 1)(x + 1)} - \dfrac{x^2 + hx + x}{(x + h + 1)(x + 1)} \\[5ex] = \dfrac{(x^2 + x + hx + h) - (x^2 + hx + x)}{(x + h + 1)(x + 1)} \\[5ex] = \dfrac{(x^2 + x + hx + h - x^2 - hx - x)}{(x + h + 1)(x + 1)} \\[5ex] = \dfrac{(x^2 + x + hx + h - x^2 - hx - x)}{(x + h + 1)(x + 1)} \\[5ex] = \dfrac{h}{(x + h + 1)(x + 1)} \\[5ex] Denominator = h \\[3ex] DQ = \dfrac{Numerator}{Denominator} \\[5ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = [f(x + h) - f(x)] \div h \\[5ex] DQ = \dfrac{h}{(x + h + 1)(x + 1)} \div \dfrac{h}{1} \\[5ex] DQ = \dfrac{h}{(x + h + 1)(x + 1)} * \dfrac{1}{h} \\[5ex] DQ = \dfrac{1}{(x + h + 1)(x + 1)}$
(12.) $f(x) = 3 + 7|x|$

$f(x) = 3 + 7|x| \\[3ex] f(x + h) = 3 + 7|x + h| \\[3ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = \dfrac{3 + 7|x + h| - (3 + 7|x|)}{h} \\[5ex] DQ = \dfrac{3 + 7|x + h| - 3 - 7|x|}{h} \\[5ex] DQ = \dfrac{7|x + h| - 7|x|}{h}$
(13.) $f(x) = \dfrac{x - 3}{x + 9}$

$f(x) = \dfrac{x - 3}{x + 9} \\[5ex] f(x + h) = \dfrac{(x + h) - 3}{(x + h) + 9} \\[5ex] f(x + h) = \dfrac{x + h - 3}{x + h + 9} \\[5ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex]$ Let us do this one at a time - piecemeal approach
Let us deal with the numerator first.

$Numerator = f(x + h) - f(x) \\[3ex] f(x + h) - f(x) = \dfrac{x + h - 3}{x + h + 9} - \dfrac{x - 3}{x + 9} \\[5ex] = \dfrac{(x + h - 3)(x + 9)}{(x + h + 9)(x + 9)} - \dfrac{(x - 3)(x + h + 9)}{(x + h + 9)(x + 9)} \\[5ex] (x + h - 3)(x + 9) = x^2 + 9x + hx + 9h - 3x - 27 = x^2 + 6x + hx + 9h - 27 \\[3ex] (x - 3)(x + h + 9) = x^2 + hx + 9x - 3x - 3h - 27 = x^2 + hx + 6x - 3h - 27 \\[3ex] = \dfrac{x^2 + 6x + hx + 9h - 27}{(x + h + 9)(x + 9)} - \dfrac{x^2 + hx + 6x - 3h - 27}{(x + h + 9)(x + 9)} \\[5ex] = \dfrac{(x^2 + 6x + hx + 9h - 27) - (x^2 + hx + 6x - 3h - 27)}{(x + h + 9)(x + 9)} \\[5ex] = \dfrac{(x^2 + 6x + hx + 9h - 27 - x^2 - hx - 6x + 3h + 27)}{(x + h + 9)(x + 9)} \\[5ex] = \dfrac{12h}{(x + h + 9)(x + 9)} \\[5ex] Denominator = h \\[3ex] DQ = \dfrac{Numerator}{Denominator} \\[5ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] DQ = [f(x + h) - f(x)] \div h \\[5ex] DQ = \dfrac{12h}{(x + h + 9)(x + 9)} \div \dfrac{h}{1} \\[5ex] DQ = \dfrac{12h}{(x + h + 9)(x + 9)} * \dfrac{1}{h} \\[5ex] DQ = \dfrac{12}{(x + h + 9)(x + 9)}$
(14.)