What are the coefficients of the function f(x) = mx^4 + nx^2 +p, if f(0)=4,f'(1)=14 and definite integral of f(x), x=0 to x=1 is 6?

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

f(0) = m*0^4 + n*0^2
+p

f(0) = p

But f(0) = 4 (from
enunciation) => p = 4

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

f'(x) = (mx^4 + nx^2
+p)'

f'(x) = 4mx^3 + 2nx

But
f'(1) = 14

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

f'(1) = 4m*1^3 +
2n*1

f'(1) = 4m + 2n

But f'(1)
= 14 => 4m + 2n = 14

We'll divide by
2:

2m +  n = 7

n = 7 – 2m
(1)

Now, we'll calculate the definite
integral:

Int f(x) dx = Int (mx^4 + nx^2 +p)dx = F(1) -
F(0)

Int (mx^4 + nx^2 +p)dx = mInt x^4dx + nInt x^2dx +
pInt dx

Int (mx^4 + nx^2 +p)dx =  m*x^5/5 + n*x^3/3 +
px

F(1) = m/5 + n/3 + p

F(0) =
0

But Int f(x)dx = 6 =>  m/5 + n/3 + p =
6

3m + 5n + 15p = 90

But p =
4=> 3m + 5n = 90 - 60

3m + 5n = 30
(2)

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

3m + 5(7 – 2m) = 30

3m+
35 – 10m = 30

-7m = -5

m =
5/7

n = 7 – 10/7

n =
39/7

The function f(x) is:
f(x) = (5x^4)/7 + (39x^2)/7 + 4