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

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