# function fFind a function f that satisfies the conditions: f " (x) = X^2 + X -1, f ' (0)=5, f (0)=10

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Since the expression of the second derivative of the original function is a quadratic, then the original function is a 4th order polynomial.

f(x) = ax^4 + bx^3 + cx^2 + dx +e

We'll substitute x by 0:

f(0) = e

But f(0) = 10 => e = 10

We'll differentiate f to get the expression of the first derivative:

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

Now, we'll substitute x by 0:

f'(0) = d

But f'(0) = 5 => d = 5

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

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

f'"(x) = 12ax^2 + 6bx + 2c (1)

But, from enunciation, we know that:

f " (x) = X^2 + X -1 (2)

We'll put (1) = (2):

12ax^2 + 6bx + 2c = X^2 + X -1

We'll put the correspondent coefficients in a relation of equality:

12a = 1

a = 1/12

6b = 1

b = 1/6

2c = -1

c = -1/2

The original function f(x) is:

f(x) = x^4/12 + x^3/6 - x^2/2 + 5x + 10