What is a, b and c?

Question

Find the values for a, b, and c so that the function
f(x) = x^3 + ax^2 + bx + c
has a critical point at (1, 5) and an inflection point at (2, 3).

Answer 1

Critical point occurs when f'(x) = 0.

Taking the derivative of f(x) you get 3x^2 +2ax + b = 0. x=1, so:
3+2a+b = 0, or
2a+b = -3

Inflection point occurs when f''(x) = 0
Take the second derivative of f(x) you get 6x + 2a = 0. x=2:
12 + 2a = 0, so a = -6.

since 2a+b = -3 --> 2(-6) +b = -3 --> b= 9.

To solve for c, just plug in either (1,5) or (2,3) into the equation.
1-6+9+c = 5
c = 1.

so a=-6, b=9, c=1.

Answer 2

Critical point is where the first derivative equals zero.
Inflection point is where the second derivative equals zero.

f' = 3x^2 + 2ax + b
f" = 6x + 2a

For f', plug in the x-value of the critical point and set =0.

3(1)^2 + 2a(1) + b = 0
3 + 2a + b = 0
2a + b = -3

For f'', plug in the x-value of the inflection point and set =0.
6(2) + 2a = 0
12 + 2a = 0
2a = -12
a = -6

Plug -6 back into the f' to find b.
2a + b = -3
2(-6) + b = -3
-12 + b = -3
b = 9

So far we have:
f(x) = x^3 - 6x^2 + 9x + c

Plug in either (1, 5) or (2, 3) to find c.

5 = (1)^3 - 6(1)^2 + 9(1) + c
5 = 1 - 6 + 9 + c
5 = 4 + c
1 = c

So the final equation is:
f(x) = x^3 - 6x^2 + 9x + 1

Answer 3

The critical points are the points where the first derivative is zero. SO:

(1) Take the first derivative;
(2) Set it to zero (Write "=0" on the right);
(3) Plug in the value "1" for x;
That will give you an equation that has "a" and "b" in it.

Next:

The inflection points are the points where the SECOND derivative is zero. SO:

(1) Take the second derivative;
(2) Set it to zero (Write "=0" on the right);
(3) Plug in the value "2" for x;
That will give you an equation that has "b" in it.

Now you have two equations in two unknowns. That should be enough to let you find the values of "a" and "b".

Next:

Go back to original function, and plug in the values you just determined for "a" and "b".

Then, write "5" in place of "f(x)", and write "1" in place of "x". This will give you an equation with just "c" in it. Solve for "c".

Answer 4

f(x) = x^3 + ax^2 + bx + c

f'(x) = 3x^2 +2ax +b
at an inflection point f'(x) = 0 so
0 = 3x^2 +2ax +b , x=2
12+4a+b=0

f"(x) = 6x + 2a
at critical point:
f"(x) = 0 = 6x + 2a
x=1
a = -3
b = 0