LOL, Calculus really
f(x+h) = 7(x+h)^2 -x-h + 3
= 7 x^2 + 14 xh + 7 h^2 -x -h + 3
now subtract f(x)
= 14 xh +7h^2 -h
now divide by h
= 14x + 7 h - 1
That is the end of the problem. To do the derivative you now look at it when h = 0
df(x)dx = 14 x - 1
Can someone please help me with this question?
If f(x) = 7x2 – x + 3 and h ≠ 0, find the following and simplify.
a.) f(x+h)
b.)( f(x+h)-f(x))/h
4 answers
f(x) = 7x^2 – x + 3
f(x+h) = 7(x+h)x^2 – (x+h) + 3
= 7x^2 + 14xh + 7h^2 - x - h + 3
( f(x+h)-f(x))/h
= (7x^2 + 14xh + 7h^2 - x - h + 3 - (7x^2 – x + 3) )/h
= (7x^2 + 14xh + 7h^2 - x - h + 3 - 7x^2 + x - 3)/h
= 14xh + 7h^2 - h)/h
= 14x + 7h - 1 , h ≠ 0
Now all you have to do is take the limit of that as h ---> 0
and you have the derivative of f(x)
f(x+h) = 7(x+h)x^2 – (x+h) + 3
= 7x^2 + 14xh + 7h^2 - x - h + 3
( f(x+h)-f(x))/h
= (7x^2 + 14xh + 7h^2 - x - h + 3 - (7x^2 – x + 3) )/h
= (7x^2 + 14xh + 7h^2 - x - h + 3 - 7x^2 + x - 3)/h
= 14xh + 7h^2 - h)/h
= 14x + 7h - 1 , h ≠ 0
Now all you have to do is take the limit of that as h ---> 0
and you have the derivative of f(x)
LOL oh well :)
Thank you guys so much! Turns out I was missing a key part.