Calculus: f'(x) = 4x^3+6x^2+2

[vv_500]

The question is: It is known that f'(x) = 4x^3+6x^2+2 and f(0)=0. Find f(x).
Are we trying to find the derivative of f(x) = 4x^3+6x^2+2. Is that what the question is asking for??
Could someone please advise me on how to approach this question? Thank you.

 

[Deleted member 4993]

The question is: It is known that f'(x) = 4x^3+6x^2+2 and f(0)=0. Find f(x).
Are we trying to find the derivative of f(x) = 4x^3+6x^2+2. Is that what the question is asking for??
Could someone please advise me on how to approach this question? Thank you.

Did you learn INTEGRATION?

 

[yoscar04]

The derivative of f(x) was provided to you. Now the question is to find f(x).
Subhotosh Khan asked you whether you have learned integration (or antiderivatives).

 

[HallsofIvy]

The question is: It is known that f'(x) = 4x^3+6x^2+2 and f(0)=0. Find f(x).
Are we trying to find the derivative of f(x) = 4x^3+6x^2+2. Is that what the question is asking for??
Could someone please advise me on how to approach this question? Thank you.

No, the problem is the opposite of "finding the derivative of f(x)= 4x^3+6x^2+ 2" because the problem clearly does NOT say "f(x)= 4x^3+6x^2+ 2", it says that "f'(x)= 4x^3+6x^2+ 2". It tells you that this is a the derivative of some function f and asks you to find f.

So how do you do that? Since you mention "the derivative" I will assume that you know how to differentiate but have not yet learned to "integrate" as Subhotosh Kahn mentions.

So how can you do this if you have not learned "integration"? Well, since you mention "find the derivative" I will assume you know that the derivative of "ax^n" is "nax^(n-1)". If you know that f'(x)= 4x^3 how do you find f? Compare nax^(n-1) and 4x^3. You should see immediately that you need na= 4 and n-1= 3. From n- 1= 3, n= 4. Then na= 4a= 4 so a= 1. The derivative of x^4 is 4x^3 so the "anti-derivative" ("integral") of 4x^3 is x^4. Now compare nax^(n-1) with 6x^2. What is "n" in this case? What is "a"? Compare nax^(n-1) with 2= 2x^0. n-1= 0 so n= 1. What is a?

 

[Steven G]

If we find the derivative of f(x) = 4x^3+6x^2+2, then we would get that f'(x) = 12x^2 + 12x. Now this has to be wrong since we were given that f'(x) = 4x^3+6x^2+2 which contradicts that f'(x) = 12x^2 + 12x.

Why would you say that f(x) = 4x^3+6x^2+2 when you were told that f'(x) = 4x^3+6x^2+2? After all f(x) and f'(x) are not generally the same.

 

[JeffM]

As everyone has written, this is a problem in integration; technically it calls for solving a separable differential equation with an initial condition.

Integrating a function means finding the family of functions such that the derivative of each function in the family equals the given function. Integration is related to differentiation the way addition is to subtraction: they are inverse operations.

In general,

[MATH]\dfrac{d}{dx}(F(x) + c) = f(x) \iff \int f(x) \ dx = F(x) + c[/MATH]
where c is an unknown constant.

Among the easiest functions to integrate are polynomials

The derivative of what function is [MATH]x^n?[/MATH]
Obviously [MATH]\dfrac{1}{n + 1} * x^{(n+1)} + c.[/MATH]
So the derivative of what function is [MATH]4x^3?[/MATH]
How about 6x²?

How about 2 = 2 * 1 = 2x⁰?