Question

use the lim f(x)-f(a)/x-a to find the derivative
x->a

f(x)=-3x^2-5x+1 and point (1,-7)

Answer 1

I know how to set it up can't factor it

Answer 2

so lim(x->a) (3x²-5x-3a²+5a)/(x-a)?

Answer 3

it has to factor because 1 will give you a zero in the numerator. so it must factor as
\[(x-1)(\text{something})\]

Answer 4

They want to use the limit definition to find the derivative. even though we know the power rule

Answer 5

ok the idea is this. you know
\[f(a)=f(1)=-7\] so
\[\frac{f(x)-f(a)}{x-a}=\frac{f(x)-(-7)}{x-1}=\frac{-x^2-5x+1+7}{x-1}=\frac{-x^2-5x+8}{x-1}\]

Answer 6

except i made a typo so let me fix it

Answer 7

\[\frac{f(x)-f(a)}{x-a}=\frac{f(x)-(-7)}{x-1}=\frac{-3x^2-5x+1+7}{x-1}=\frac{-3x^2-5x+8}{x-1}\]

Answer 8

right I got up to that part but stuck here

Answer 9

you know for a fact that if you replace x by 1 in the numerator you will get zero. that has to be because you subtracted f(1)

Answer 10

this means it has to factor, and it has to factor as
\[(x-1)(\text{whatever})\]

Answer 11

3x²-5x-3a²+5a factors to 3(x²-a²)-5(x-a)=(x-a)(3(x+a)-5)

Answer 12

so factor out an x -1 and see what the other term has to be. it has to be
\[\frac{(x-1)(-3x-8)}{x-1}\]

Answer 13

now cancel the factor of x - 1 to get
\[-3x-8\] and finally replace x by 1 to get
\[-11\]

Answer 14

oh no, it is right

Answer 15

could you show me how you factored the top

Answer 16

sure, but lets make sure one thing is clear. you have the numerator of
\[-3x^2-5x+8\] right? and how did you get it? you took
\[-3x^2-5x+1\] replaced x by 1 and got -7, then subtracted it off to get
\[-3x^2-5x+8\]

Answer 17

which means for sure if you replace x by 1 you have to get 0

Answer 18

so you already have one zero of this polynomial, it is 1. which means it MUST factor as
\[(x-1)(\text{something})\]

Answer 19

your only job now is to figure out what the something is. so it is in fact much easier than factoring in general because you already know what one of the factors must be. it must be (x- 1)

Answer 20

so ask yourself if
\[-3x^2-5x+8=(x-1)(\text{something})\] what must the something be? well the first term must be -3x
and the second term must be -8 or else it wont work

Answer 21

so is it safe to assume that the denominator is going to be a factor (in this x-1)

Answer 22

it is safe to assume that if you replace x by 1 you get zero so it MUST factor that way, yes

Answer 23

that is what the "factor" theorem says. if you have a zero of a polynomial you can factor it out. so don't go looking to factor as in "gee what can it be" you already know one of the factors, you only have to find the other one

Answer 24

and certainly it will cancel with the denoninator. of course you already know the derivative is
\[f'(x)=-6x-5\] and if you replace x by 1 here you will get
\[f'(1)=-11\]

Answer 25

I get the rest just got stuck on the factoring thanks a lot