# f(x)=ax^4+bx^2+c What are a,b,c if f(0)=2 , f'(1)=26 and the definite integral of f(x) for x=0 to x=1 is 4?

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### 1 Answer

We'll determine f(0).

To get f(0), we'll replace x by 0 in the expression of f(x):

f(0) = a*0^4 + b*0^2 +c

f(0) = c

But f(0) = 2 (from enunciation) => c = 2

Now, we'll calculate f'(x):

f'(x) = (ax^4 + bx^2 +c)'

f'(x) = 4ax^3 + 2bx

But f'(1) = 26

We'll calculate f'(1) substituting x by 1 in the expression of the first derivative:

f'(1) = 4a*1^3 + 2b*1

f'(1) = 4a + 2b

But f'(1) = 26 => 4a + 2b = 26

We'll divide by 2:

2a + b = 13

b = 13 - 2a (1)

We'll calculate the definite integral:

Int f(x) dx = Int (ax^4 + bx^2 +c)dx = F(1) - F(0)

Int (ax^4 + bx^2 +c)dx = aInt x^4dx + bInt x^2dx + cInt dx

Int (ax^4 + bx^2 +c)dx = a*x^5/5 + b*x^3/3 + cx

F(1) = a/5 + b/3 + c

F(0) = 0

But Int f(x)dx = 4 => a/5 + b/3 + c = 4

3a + 5b + 15c = 60

But c = 2=> 3a + 5b = 60 - 30

3a + 5b = 30 (2)

We'll substitute (1) in (2):

3a + 5(13 - 2a) = 30

3a + 65 - 10a = 30

-7a = -35

a = 5

b = 13 - 10

b = 3

The function f(x) is: f(x) = 5x^4 + 3x^2 + 2