Difference Quotient

[nycmathguy]

Find the difference quotient
and simplify your answer.

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)

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

I don't think this can be simplified further.

You say?

 

[MathLover1]

Find the difference quotient
and simplify your answer.

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)

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

I don't think this can be simplified further.

You say?

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

 

[nycmathguy]

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

Understood.