# Difference Quotient Part 2

Discussion in 'Algebra' started by nycmathguy, Jul 4, 2021.

1. ### nycmathguy

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I want to continue with this question posted on 7/3/21.

Find the difference quotient

f(x) = x^(2/3) + 1

[f(x) - f(8)]/(x - 8), where x cannot be 8.

I will find f(8) first.

Note: x^(2/3) = cuberoot{x^2}.

cuberoot{x^2} + 1

cuberoot{8^2} + 1

cuberoot{64} + 1

16 + 1 = 17

[cuberoot{x^2} - 17]/( x - 8)

Rationalize the numerator.

Numerator:

[cuberoot{x^2} - 17][cuberoot{x^2} + 17]

Denominator:

(x - 8)[cuberoot{x^2} + 17]

Let cr = cube root

[cr{x^2} - 17][cr{x^2} + 17]/(x - 8)[cr{x^2} + 17]

Simplify the numerator. Leave denominator alone.

Note: (cr{x^2})^2 = x^(4/3)

Note: (-17)(17) = -289

Finally, I came up with this:

[x^(4/3) - 289]/(x - 8)

I don't think this can be simplified further.

You say?

nycmathguy, Jul 4, 2021

2. ### MathLover1

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you are using wrong formula

f(x) = x^(2/3) + 1

[f(x+h) - f(x)]/ h, where h cannot be 0

first find f(x+h)

f(x+h) = (x+h)^(2/3) + 1
f(x+h) = (x+h)^(2/3) + 1

then substitute in [f(x+h) - f(x)]/ h both f(x) and f(x+h)

[ (x+h)^(2/3) + 1 - x^(2/3) + 1]/ h

[ (h + x)^(2/3) - x^(2/3) + 2]/ h

or
(h + x)^(2/3) /h- x^(2/3)/h+ 2/ h

MathLover1, Jul 4, 2021
nycmathguy likes this.

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