Find f(x+h) -f(x) all over h...given f(x)= sqrt of all over 2x +3 . Rationalize the numerator if possible.
3 answers
it doesnt say what f(x) is the sqrt of?
If you use parentheses, you would be able to see the expression and give it a good check also. For example
(f(x+h)-f(x))/h where f(x) = sqrt(2x+3)
Be very careful when you interpret fractions. Even the expression does not show parentheses, you need to insert parentheses around the numerator and the denominator, like
(x-4)/(x²-16) and not x-4/x²-16.
Back to the origina question:
(f(x+h)-f(x))/h where f(x) = sqrt(2x+3)
The simple answer would be:
(sqrt(2(x+h)+3)-sqrt(2x+3))/h
=(sqrt((2(x+h)+3)/h²)-sqrt((2x+3)/h²))
This is about as far as you can go.
If you are eventually taking the limit h->0, then you can further simplify using series expansions.
(f(x+h)-f(x))/h where f(x) = sqrt(2x+3)
Be very careful when you interpret fractions. Even the expression does not show parentheses, you need to insert parentheses around the numerator and the denominator, like
(x-4)/(x²-16) and not x-4/x²-16.
Back to the origina question:
(f(x+h)-f(x))/h where f(x) = sqrt(2x+3)
The simple answer would be:
(sqrt(2(x+h)+3)-sqrt(2x+3))/h
=(sqrt((2(x+h)+3)/h²)-sqrt((2x+3)/h²))
This is about as far as you can go.
If you are eventually taking the limit h->0, then you can further simplify using series expansions.
Actually, you can rationalize the numerator as follows by multiplying both the numerator and denominator by: (sqrt(2(x+h)+3)+sqrt(2x+3))
(sqrt(2(x+h)+3)-sqrt(2x+3))/h
=(2(x+h)+3)-(2x+3)/(h(sqrt(2(x+h)+3)+sqrt(2x+3)))
=(2h)/(h(sqrt(2(x+h)+3)+sqrt(2x+3)))
=2/(sqrt(2(x+h)+3)+sqrt(2x+3))
(sqrt(2(x+h)+3)-sqrt(2x+3))/h
=(2(x+h)+3)-(2x+3)/(h(sqrt(2(x+h)+3)+sqrt(2x+3)))
=(2h)/(h(sqrt(2(x+h)+3)+sqrt(2x+3)))
=2/(sqrt(2(x+h)+3)+sqrt(2x+3))