Question

the function f is given by the formula
f(x)= (3x^3-5x^2+8x-20)/(x-2) when x<2
and by f(x) = -5x^2-4x+a when x greater then or equal to 2.
What value must be chosen for a in order to make this function continuous at 2?

I have tried to work this out multiple times and I keep getting 0. How do you do this correctly?

Answer 1

Can you factor the numerator of the first definition of f(x) (i.e. the part for x<2)?

Answer 2

for x<2, f(x) = (3x^3-5x^2+8x-20)/(x-2)
= (3x^2 + x + 10)(x-2) / (x-2)
= (3x^2 + x + 10)

Answer 3

for x = 2, you use the 2nd definition of f(x), f(x) = -5x^2 - 4x +a
f(2) = -5(2)^2 - 4(2) + a
f(2) = -20-8+a
f(2) = -28 + a

Answer 4

So going back to the first definition of f(x), but using the version after we factored and cancelled annoying (x-2) denominator, you can plug in f(2) and get the value of f(2). For it to be continuous at x=2, the value of f(2) from the first expression must match the f(2) = -28 +a from the 2nd expression, so you can solve for a.

Answer 5

how were you able to factor out the (x-2) from the first one? This probably seems like a silly question

Answer 6

I had a hunch that you would have to be able to factor it out to be able to cancel it with the denominator. But honestly, I am not good at factoring cubes like that... I used Wolfram Alpha to find alternate forms, and I found the factoring there. (Although I did work it through to be sure it actually worked... :) )

Answer 7

thank you for your help!! :)

Answer 8

Look for those "hints" with the denominator terms that are like (x-5) or whatever... they usually are the same point at which the function definition has to change, so it's f(x) one way for x<5, and a different f(x) for x>5

Answer 9

glad to help :)