Find a function f that satisfies the conditions: f " (x) = X^2 + X -1, f ' (0)=5, f (0)=10
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f " (x) = X^2 + X -1, f ' (0)=5, f (0)=10
First we need to determine f'(x) by calculating the integral of f''(x)
We know that:
f'(x) = intg f''(x)
==> f'(x) = intg (x^2 + x -1) dx
= x^3/3 + x^2/2 - x + C
==> f'(x) = (1/3)x^2 + (1/2)x^2 - x + C
Given that:
f'(0) = 5
==> f'(x) = (1/3)x^3 + (1/2)x^2 - x + 5
Now let us calculate f(x)"
We know that:
f(x) = intg f'(x)
= intg ((1/3)x^3 + (1/2)x^2 - x + 5) dx
= (1/3)x^4/4 + (1/2)x^3/3 - x^2/2 + 5x + C
= (1/12)x^4 + (1/6)x^3 - (1/2)x^2 + 5x + C
Given that:
f(0) = 10 ==> c = 10
==> f(x) = (1/12)x^4 + (1/6)x^3 - (1/2)x^2 + 5x + 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
To find the the function for which ,
f " (x) = X^2 + X -1, f ' (0)=5, f (0)=10.
Sinxe f"(x) = x^2+x-1, f'(x) is the integral of f"(x) dx.
Therefore f'(x) = Int (x^2+x-1)dx = x^3+x^2/2 -x +C.
Now we apply the given conditions: f'(0) = 5.
So f'(0) = 0^3/3+0^2/2 -0 +C = 5.. So C = 5.
Therefore f'(x) = x^3/3+x^2/2 -x+5.
So f(x) = Int f'(x) dx = Int (x^3/3+x^2/2 -x+5) dx = x^4/(4*3) +x^3/(2*3) -x^2/2 +5x + C2
f(x) = x^4/12+x^3/6-x^2/3+5x+C2
But f(0) =10 = 0 + C2. So C2 = 10.
Therefore f(x) = x^4/12+x^3/6-x^2/3+5x +10.
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