Question

limit definition of the derivative to find f ' (x) for f(x)= x2 + 3x + 1

Answer 1

first can you tell me what f(x+h) is in this case?

Answer 2

No idea ...

Answer 3

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

f(x+h) = (x+h)^2 + 3(x+h) + 1 ... replace all 'x' terms with 'x+h'

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

Answer 4

now you have this, use this and plug it into

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

Answer 5

why do we use f(x+h) ?

Answer 6

and what am I plugging into [f(x+h) - f(x)]/h ?

Answer 7

sry the site is being slow

Answer 8

we use f(x+h) because if you drew this out, you'll have a secant line at (x,f(x)) and (x+h, f(x+h))

a drawing might help see this

now imagine dragging one point closer to another fixed point....you'll have that secant line slowly turn into a tangent line...which is exactly what the derivative helps us find

Answer 9

anyways

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

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

[x^2 + 2xh + h^2 + 3x + 3h + 1 - x^2 - 3x - 1]/h

[2xh + h^2 + 3h]/h

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

2x + h + 3

Answer 10

Then you take the limit as h ---> 0

ie you plug in h = 0 to get

2x + h + 3

2x + 0 + 3

2x + 3

Answer 11

so the derivative function is

f ' (x) = 2x + 3

Answer 12

okay thank you so much ! that makes a lot more sense especially when I graphed it

Answer 13

yeah a picture says 1000 words often, so I'm glad it's clicking