How do you solve this tough pre-Calculus problem?

Question

Let f: R-R be defined by f(x)=3x^2-2x+5. Find lim (f(x)-f(a))/(x-a) when x-a.

Answer 1

lim (f(x)-f(a))/(x-a) =f ' (a)

This means that you first derivate f(x) and you obtain 6x-2, and then you put a instead x.

So the result is : 6a-2



(3x^2)'=3*2*x
(-2x)'= -2*1= -2
5'=0

Answer 2

approach a million: notice that it somewhat is only yet in any different case of writing the by-product. end that it somewhat is f'(x) evaluated at a, this is 6a-2 approach 2: considering the fact that direct substitution promises 0/0, use L'hopitals rule and locate that [x?a]lim (f(x)-f(a))/(x-a) = [x?a]lim f'(x)/a million = f'(a) = 6a-2 (in truth comparable to above, yet using a rule fairly than the similarity to the version quotient) approach 3: Direct attitude: [x?a]lim (3x² - 2x + 5 - (3a² - 2a + 5))/(x-a) Simplify: [x?a]lim (3x² - 3a² - 2x + 2a)/(x-a) chop up the numerator during two fractions: [x?a]lim (3x² - 3a²)/(x-a) - (2x - 2a)/(x-a) factor the numerators: [x?a]lim 3(x+a)(x - a)/(x-a) - 2(x-a)/(x-a) Cancel: [x?a]lim 3(x+a) - 2 evaluate: 3(a+a)-2 6a-2

Answer 3

Sorry, I don't know what f: R-R means. Also it appears your sentence got cut off at the end.

But (f(x) - f(a))/(x-a) =
(3x^2-2x+5 -3a^2 + 2a - 5)/(x-a)
=(3x^2-2x -3a^2 +2a)/(x-a)
=[3(x^2-a^2) - 2(x -a)]/(x-a)
= 3(x+a) - 2
when a --> 0 we get 3x - 2 as the limit.

Answer 4

Good luck with that. I had calculus but happily forgot it.

Answer 5

...when x - a?????