Difference Quotient

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 2, Section 2.1

Find the difference quotient of f; that is find
[f(x + h) - f(x)]/h, h cannot be 0, for each function. Be sure to simplify.

85. f(x) = 1/x

86. f(x) = 1/(x + 3)

[Math Tutor]

Both problems use the same trick: combine the two fractions into one first, then divide by h at the very end.

85. f(x) = 1/x

[f(x+h) - f(x)]/h = [1/(x+h) - 1/x] / h

Common denominator x(x+h):
1/(x+h) - 1/x = [x - (x+h)] / [x(x+h)] = -h / [x(x+h)]

Now divide by h (multiply by 1/h); the h's cancel:
= -1 / [x(x+h)], h =/= 0

86. f(x) = 1/(x+3)

Same idea, common denominator (x+3)(x+h+3):
1/(x+h+3) - 1/(x+3) = [(x+3) - (x+h+3)] / [(x+3)(x+h+3)] = -h / [(x+3)(x+h+3)]

Divide by h:
= -1 / [(x+3)(x+h+3)], h =/= 0

Notice 86 is just 85 shifted 3 units, so its answer is 85's answer with x replaced by x+3.

[Eigenvalue]

[tex]\frac{(1/(x+h))-(1/x)}{h}[/tex]

=[tex]\frac{x-(x+h)}{h(x)(x+h)}[/tex]

=[tex]\frac{-h}{h(x)(x+h)}[/tex]

=[tex]\frac{-1}{x(x+h)}[/tex]

[Eigenvalue]

[tex]\frac{(1/x+h+3)-(1/x+3)}{h}[/tex]

=[tex]\frac{x+3-(x+h+3)}{h(x+3)(x+h+3)}[/tex]

=[tex]\frac{-h}{h(x+3)(x+h+3)}[/tex]

=[tex]\frac{-1}{(x+3)(x+h+3)}[/tex]

[nycmath]

Both problems use the same trick: combine the two fractions into one first, then divide by h at the very end.

85. f(x) = 1/x

[f(x+h) - f(x)]/h = [1/(x+h) - 1/x] / h

Common denominator x(x+h):
1/(x+h) - 1/x = [x - (x+h)] / [x(x+h)] = -h / [x(x+h)]

Now divide by h (multiply by 1/h); the h's cancel:
= -1 / [x(x+h)], h =/= 0

86. f(x) = 1/(x+3)

Same idea, common denominator (x+3)(x+h+3):
1/(x+h+3) - 1/(x+3) = [(x+3) - (x+h+3)] / [(x+3)(x+h+3)] = -h / [(x+3)(x+h+3)]

Divide by h:
= -1 / [(x+3)(x+h+3)], h =/= 0

Notice 86 is just 85 shifted 3 units, so its answer is 85's answer with x replaced by x+3.

Wow! Impressive. You outdid yourself again.

[nycmath]

[tex]\frac{(1/(x+h))-(1/x)}{h}[/tex]

=[tex]\frac{x-(x+h)}{h(x)(x+h)}[/tex]

=[tex]\frac{-h}{h(x)(x+h)}[/tex]

=[tex]\frac{-1}{x(x+h)}[/tex]

Correct. Good job!

[nycmath]

[tex]\frac{(1/x+h+3)-(1/x+3)}{h}[/tex]

=[tex]\frac{x+3-(x+h+3)}{h(x+3)(x+h+3)}[/tex]

=[tex]\frac{-h}{h(x+3)(x+h+3)}[/tex]

=[tex]\frac{-1}{(x+3)(x+h+3)}[/tex]

You are cooking with gas.