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?

justaguide | College Teacher | (Level 2) Distinguished Educator

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

f(0) = 4

=> m*0^4 + n*0^2 +p = 4

=> p = 4

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

f'(1) = 14

=> 4*m*1 + 2*n*1 = 14

=> 4m + 2n = 14

=> 2m + n = 7 ...(1)

The integral of f(x) is

m*x^5 / 5 + n*x^3 / 3 + px

Between the limits x = 0 and x = 1, the integral is equal to

m*1^5 / 5 + n*1^3 / 3 + p*1 - m*0^5 / 5 + n*0^3 / 3 - p*0

=> m/5 + n/3 + 4

This is equal to 6

=> m/5 + n/3  + 4 = 6

=> 3m + 5n = 30 ...(2)

5*(1) - (2)

=> 10m + 5n - 3m - 5n = 35 - 30

=> 7m = 5

m = 5/7

n = 39/7

The coefficients of the function are p = 4, m = 5/7 and n = 39/7

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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