If f'(x) = 2x^2 - 6x + 7, find f(x) if f(0) = -8.
- print Print
- list Cite
Expert Answers
calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We are given the derivative of f(x) as f'(x) = 2x^2 - 6x + 7. We have to find f(x) if f(0) = -8.
Now Int [2x^2 - 6x + 7]
=> Int [2x^2] - Int [6x] + Int [7] + C
=> 2*x^3/3 - 6x^2/2 + 7x + C
=> (2/3)x^3 - 3x^2 + 7x + C
So f(0) = (2/3)*0^3 - 3*0^2 + 7*0 + C = -8
=> C = -8
Therefore f(x) = (2/3)x^3 - 3x^2 + 7x - 8
The required function is f(x) = (2/3)x^3 - 3x^2 + 7x - 8.
Related Questions
- x^2 + 2x -8 > 0 find x values.x^2 + 2x -8 > 0 find x values.
- 1 Educator Answer
- Find f. f''(x) = 6 + 6x + (24x^2) , f(0) = 5, f(1) = 14
- 1 Educator Answer
- find f(x) Knowing f'(x) = 2x^2 - 6x + 7, find f(x) if f(0) = -8.
- 1 Educator Answer
- find f(x) given that f'''(x)=cos x, f(0)=8, f'(0)=4 and f''(0)=9.
- 1 Educator Answer
- if f(x)=2x-3 and fg(x)=2x+1,find g(x)
- 1 Educator Answer
calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
f'(x) = 2x^2 - 6x + 7 find f(x) if f(0) = -8.
We are given the first derivative of the function and we need to determine the function.
Then we know that:
f(x) = integral of f'(x).
==> f(x) = intg (2x^2 - 6x + 7) dx
==> f(x) = intg (2x^2) dx - intg (6x) dx + intg 7 dx
==> f(x) = 2x^3/3 - 6x^2/2 + 7x + C
==> f(x) = (2/3)x^3 - 3x^2 + 7x + C
But we are given that f(0) = -8.
Then we will substitute with x= 0.
==> f(0) = 0 - 0 + 0 + C = -8
==> C = -8.
Then, we will substitute with c= -8 into the function.
==> f(x) = (2/3)x^2 - 3x^2 + 7x - 8
If f'(x) = 2x^2 - 6x + 7 find f(x) if f(0) = -8..
We know that f(x) = Int f'(x) dx.
So if f'(x) = 2x^2-6x+7 , then f(x) = Int (2x^2-6x+7).dx
Int (2x^2-6x+7) dx = Int (2x^2-Int(6xdx)+ Int 7dx + C.
Int (2x^2-6x+7) dx = 2*(1/3)x^3 -6*(1/2)x^2+ 7x+C.
Int (2x^2-6x+7) dx = (2/3)x^3-3x^2+7x +C.
Therefore f(x) = (2/3)x^3-3x^2+7x +C.
Put x= 0 in f(x) = (2/3)x^3-3x^2+7x +C and we get:
f(0) = 0+C. But f(0) = -8.
So c= -8. So rewite with c = -8 in f(x) = (2/3)x^3-3x^2+7x +C :
f(x) = (2/3)x^3-3x^2+7x -8.
If f'(x) = 2x^2 - 6x + 7, that means that the function f(x) is a polynomial of 3rd order:
f(x) = ax^3 + bx^2 + cx + d
If f(0) = -8, we'll substitute x by 0 and we'll get:
f(0) = d
d = -8
f'(x) = 3ax^2 + 2bx + c
We'll compare f'(x) with the given f'(x) and we'll get:
3a = 2
a = 2/3
2b = -6
b = -3
c = 7
The function f(x) = 2x^3/3 - 3x^2 + 7x - 8.
Student Answers