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

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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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