Question

@asnaseer

Answer 1

this is a question like the other one

Answer 2

do you know how to do a question like this?

Answer 3

ok - this is /slightly/ different....

Answer 4

oh....

Answer 5

we are given:\[f(x)=4.5x^2-3x+2\]and are asked to use the alternate limit definition which states that:\[f'(a)=\lim_{x\rightarrow a}{\frac{f(x)-f(a)}{x-a}}\]

Answer 6

so now we first calculate f(a) from our definition to get:\[f(a)=4.5a^2-3a+2\]therefore:\[f(x)-f(a)=4.5x^2-3x+2-(4.5a^2-3a+2)\]\[=4.5x^2-3x+2-4.5a^2+3a-2\]\[=4.5(x^2-a^2)-3(x-a)\]understand so far?

Answer 7

ohhh yeah its a little diff

Answer 8

but yeah i comprehend

Answer 9

the "trick" here again is to avoid dividing by zero. As x tends to "a" (x-a) tends to zero

Answer 10

so we need to get an (x-a) factor in the numerator so that it cancels out with the one in the denominator

Answer 11

so, we have:\[f(x)-f(a)=4.5(x^2-a^2)-3(x-a)\]therefore:\[f'(a)=\lim_{x\rightarrow a}{\frac{f(x)-f(a)}{x-a}}=\lim_{x\rightarrow a}{\frac{4.5(x^2-a^2)-3(x-a)}{x-a}}\]agreed?

Answer 12

okk

Answer 13

now notice we have a term in the numerator that involves \((x^2-a^2)\). Do you know how to factor this? (It is a difference of two squares which has a well known factoring)

Answer 14

yes! (x-a)(x+a)

Answer 15

sorry my internet was slow

Answer 16

good, so we get:\[f'(a)=\lim_{x\rightarrow a}{\frac{4.5(x^2-a^2)-3(x-a)}{x-a}}\]\[=\lim_{x\rightarrow a}{\frac{4.5(x-a)(x+a)-3(x-a)}{x-a}}\]\[=\lim_{x\rightarrow a}{\frac{(x-a)(4.5(x+a)-3)}{x-a}}\]\[=\lim_{x\rightarrow a}{4.5(x+a)-3}\]Now we can actually take the limit by just setting x=a to finally get:\[f'(a)=4.5(a+a)-3=9a-3\]

Answer 17

from this you should be able to calculate f'(2)

Answer 18

15 (:

Answer 19

NOTE: This matches our result from the previous question which said \(f'(x)=9x-3\)

Answer 20

15 is correct! :)

Answer 21

This is just an alternate means of working out f'

Answer 22

thank you so much!!! your steps are extremely clear and i understand how to do it(:

Answer 23

yw - I'm glad I was able to help :)