# Using integral calculus to find the equation of the quartic

#### Jaimee

1. Homework Statement
The question states Use integral calculus to find the euation of the quartic that has (1,23) and (3, 15) and a y-intercept of 24.

2. Homework Equations
The previous part of the question was A quartic has stationary points of inflection at x=1 and x=3. Explain why f"(x)=k(x-1)(x-3) where k doesn't =0

3. The Attempt at a Solution
Part A:

A quartic has a non-stationary points of inflection at x=1 and x=3. Explain why f”(x)=k(x-1)(x-3) k≠0

A quartics points of inflection can be found by equating its second derivative to zero.

as f”(x)=k(x-1)(x-3) if f”(x)=0

0=k(x-1)(x-3)

if k≠0

0=(x-1)(x-3)

and hence x=1 and x=3

Part B:

Use integral calculus to find the equation of the quartic that has (1,23) and (3,15) and a y-intercept of 24.

f”(x)=k(x-1)(x-3)

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#### Big Red Car

Hi Jaimee, have you tried integrating the second derivative? Or are you unsure how to do the integration?

#### Jaimee

Currently i have this
f”(x)=k(x-1)(x-3)

∫f”(x)=∫k(x-1)(x-3)

f’(x)=k∫(x−3)(x−1)dx

f’(x)=((k(x−3)^2 (x))/3)+C

but f(x)=24 x=0

f’(0)=0

0=((k(0−3)^2 (0))/3)+C

0=0+C

C=0
therefore f'(x)=((k(x−3)^2 (x))/3)

#### Mark44

Mentor
Thread moved -- if the title includes "calculus" the thread doesn't belong in the Precalculus section.

#### Big Red Car

Currently i have this
f”(x)=k(x-1)(x-3)

∫f”(x)=∫k(x-1)(x-3)

f’(x)=k∫(x−3)(x−1)dx

f’(x)=((k(x−3)^2 (x))/3)+C

but f(x)=24 x=0

f’(0)=0

0=((k(0−3)^2 (0))/3)+C

0=0+C

C=0
therefore f'(x)=((k(x−3)^2 (x))/3)
I'm not sure I follow the third line: f’(x)=((k(x−3)^2 (x))/3)+C
Maybe try expanding (x-3)(x-1) before integrating. Also you want a quartic (order 4) so does that mean your answer can be found with the first derivative? Or do you need to do an extra step?

EDIT: I mean the 4th line not the third. In future its good practice to number your equations, it will help people help you! :).

#### Mark44

Mentor
Currently i have this
f”(x)=k(x-1)(x-3)

∫f”(x)=∫k(x-1)(x-3)

f’(x)=k∫(x−3)(x−1)dx

f’(x)=((k(x−3)^2 (x))/3)+C
As already noted, the line above is wrong. I can't tell what you did. You should expand (x - 3)(x - 1) before atempting to integrate.
Jaimee said:
but f(x)=24 x=0
From the given information about the y-intercept, f(0) = 24, not f(x) = 24.

Starting with f''(x), you need to integrate twice to get f(x). At each integration there will be a constant of integration.
Use integral calculus to find the equation of the quartic that has (1,23) and (3,15) and a y-intercept of 24.
When you get to a formula for f(x), there should be two constants of integration. Use the fact that f(1) = 23, f(3) = 15, and f(0) = 24 to determine those constants of integration and the constant k.

Your function should be a fourth-degree polynomial with all coefficients determined.