help with calculus?

Question

what is the cubic function f(x) that satisfies the following conditions?

a) when it is divided by (x-2)² the remainder is 2x+1.

b) it has a relative extreme value of 2 at x=1.

i prefer an explanation over an answer. thanks!

Answer 1

Well, the first condition (a) implies that:

f(x) = (x-2)^2(ax+b) + 2x+1

Where (ax+b) is the factor that makes f(x) a cubic.

The second condition implies that f'(1) = 0, and that f(1) = 2.

So, if we expand f(x) (as shown above), we get:
f(x) = ax^3 + (b-4a)x^2 + (4a-4b+2)x + (4b+1)

f'(x) = 3ax^2 + 2(b-4a)x + (4a-4b+2)

From setting f(1) = 2:
2 = a+(b-4a)+(4a-4b+2)+(4b+1)
a+b=-1

From setting f'(1) = 0:
0 = 3a + 2(b-4a) + (4a-4b+2)
-a-2b=-2
a+2b=2

Solving (say, rearrange the first to (a=-1-b), and plugging into the second):
(-1-b)+2b = 2 --> b=3.

and so a = -1-b = -4.

Therefore f(x) = ax^3 + (b-4a)x^2 + (4a-4b+2)x + (4b+1)
f(x) = -4x^3 + (3+16)x^2 + (-16-12+2)x+(-12+1)
= -4x^3 + 19x^2 - 26x - 11.

Answer 2

So you have some function in the form of:
f(x) = ax^3 + bx^2 + cx + d

This is divided by (x-2)^2 to give a remainder of 2x+1, so this means the function is some multiple of (x-2)^2, plus 2x+1. Let this multiple be (mx + n) for some constants m and n. (This is because we need one x term to get it into a cubic.)

This means
f(x) = (mx + n)(x-2)^2 + 2x+1

Expand and simplify this to get:
(mx + n)(x-2)^2 + 2x+1
(mx + n)(x^2 - 4x + 4) + 2x+1
mx^3 + nx^2 - 4mx^2 - 4nx + 4mx + 4n + (2x+1)
mx^3 + (n - 4m)x^2 + (-4n + 4m + 2)x + (4n + 1)
So this gives you a general form of f(x).

Part b) basically tells us that f(1) = 2, and f ' (1) = 0. You can use this to find the value sof m and n.

Answer 3

y== a(x-p)((x-2)^2+2x+1
y´= a((x-2)^2+2(x-p)(x-2))+2
a tx=1 y´=0 so a(1-2+2p)+2=0
and y(1)=2 so a(1-p)+3=2
You can solve for a and p