Given f''(x) = 4x - 3
and f'( -3) =2 and f( -3)=-1.
Find f'(x) =______
and find f( 1) =_____
y '' = 4x+3
y' = 2x^2 + 3x + C
when x = -3, y' = 2
2 = 18 - 9 + C
C = -7
so y' = 2x^2 + 3x - 7
then y = (2/3)x^3 + (3/2)x^2 - 7x + K
when x = -3, y = 1
take it from there by solving for K
I don't know why but the answer is wrong.
To find f'(x), we need to integrate f''(x). Let's start with the given equation:
f''(x) = 4x - 3
To integrate f''(x), we need to find the antiderivative of 4x - 3. The antiderivative of 4x is 2x^2, and the antiderivative of -3 is -3x. So we have:
f'(x) = 2x^2 - 3x + C
Where C is the constant of integration.
To determine the value of C, we can find f'( -3) using the given information:
f'( -3) = 2
Plugging -3 into our equation for f'(x), we get:
2 = 2(-3)^2 - 3(-3) + C
2 = 18 + 9 + C
2 = 27 + C
C = 2 - 27
C = -25
Now we can substitute C back into our equation for f'(x):
f'(x) = 2x^2 - 3x - 25
To find f(1), we need to integrate f'(x) to get f(x). Integrating 2x^2 - 3x - 25, we get:
f(x) = (2/3)x^3 - (3/2)x^2 - 25x + D
Where D is the constant of integration.
To determine the value of D, we can find f( -3) using the given information:
f( -3) = -1
Plugging -3 into our equation for f(x), we get:
-1 = (2/3)(-3)^3 - (3/2)(-3)^2 - 25(-3) + D
-1 = (-2) - (9/2) + 75 + D
-1 = -9/2 - 18/2 + 150/2 + D
-1 = ( -9 - 18 + 150)/2 + D
-1 = 123/2 + D
D = -1 - 123/2
D = -2/2 - 123/2
D = -125/2
Now we can substitute D back into our equation for f(x):
f(x) = (2/3)x^3 - (3/2)x^2 - 25x - (125/2)
Therefore, f'(x) = 2x^2 - 3x - 25
And f(1) = (2/3)(1)^3 - (3/2)(1)^2 - 25(1) - (125/2) = 2/3 - 3/2 - 25 - 125/2 = -189/6 = -63/2