1. ## Differentiable function

Let $f$ be a differentiable function, defined for all real numbers $x$, with the following properties. Find $f(x)$. Show your work.
- $f '(x)=ax^2 + bx$
- $f '(1)=6$ and $f ''(1)=18$
. $2$
- ∫ $f(x)dx=18$
. $1$

I am completely lost on this question. I assume that I have to take the anti derivative of the first given property by using the second given to solve for the values of a and b, but I'm not sure how to do it.

2. Originally Posted by DarkestEvil
Let $f$ be a differentiable function, defined for all real numbers $x$, with the following properties. Find $f(x)$. Show your work.
- $f '(x)=ax^2 + bx$
- $f '(1)=6$ and $f ''(1)=18$
. $2$
- ∫ $f(x)dx=18$
. $1$

I am completely lost on this question. I assume that I have to take the anti derivative of the first given property by using the second given to solve for the values of a and b, but I'm not sure how to do it.
$f'(x) = ax^2 + bx$

$f'(1) = a(1^2) + b(1) = a+b = 6$

$f''(x) = 2ax + b$

$f''(1) = 2a(1) + b = 2a + b = 18$

you have a system of equations in $a$ and $b$ ... solve for both.

once you get $a$ and $b$, integrate $f'(x)$ to get $f(x)$ with its constant of integration ... then use the information given by the definite integral.

3. Originally Posted by DarkestEvil
Let $f$ be a differentiable function, defined for all real numbers $x$, with the following properties. Find $f(x)$. Show your work.
- $f '(x)=ax^2 + bx$ Mr F says: Therefore ${\color{red}f(x) = \frac{a}{3} x^3 + \frac{b}{2} x^2 + c}$.

- $f '(1)=6$ Mr F says: Therefore ${\color{red}6 = a + b}$ .... (1)

and $f ''(1)=18$ Mr F says: ${\color{red}f''(x) = 2ax + b}$. Therefore ${\color{red}18 = 2a + b}$ .... (2)
. $2$
- ∫ $f(x)dx=18$
. $1$
Mr F says: Therefore ${\color{red}18 = \left[\frac{a}{12} x^4 + \frac{b}{6} x^3 + cx + d\right]_1^2}$ .... (3)

I am completely lost on this question. I assume that I have to take the anti derivative of the first given property by using the second given to solve for the values of a and b, but I'm not sure how to do it.
Solve equations (1) and (2) simultaneously for a and b. Equation (3) gives a relationship between c and d but without further information exact values cannot be found.

Edit: Re: "Equation (3) gives ....." etc. My mistake, d obviously cancels so getting c is simple.

4. Wow that makes so much more sense now using system of equations. I found that $a=12$ and $b=-6$. So now I'm guessing I take the anti derivative of $f'(x)$, making it $f(x)=(ax^3/3) + (bx^2/2) + C$. But wouldn't I have to solve for C to be able to find $f(x)$?

5. Originally Posted by DarkestEvil
Wow that makes so much more sense now using system of equations. I found that $a=12$ and $b=-6$. So now I'm guessing I take the anti derivative of $f'(x)$, making it $f(x)=(ax^3/3) + (bx^2/2) + C$. But wouldn't I have to solve for C to be able to find $f(x)$?
that is correct ... use the given definite integral information to solve for C

6. Hmmm... how do I find C without an x value? Do I use $f'(x)$ to find the value of x by plugging in a and b? And even so, how do I find C when I have both unknowns $f(x)$ and C in the same equation of $f(x)=(ax^3/3)+(bx^2/2)+C$?

7. Originally Posted by DarkestEvil
Hmmm... how do I find C without an x value? Do I use $f'(x)$ to find the value of x by plugging in a and b? And even so, how do I find C when I have both unknowns $f(x)$ and C in the same equation of $f(x)=(ax^3/3)+(bx^2/2)+C$?
evaluate and get an equation in terms of C, then solve for C ...

$\int_1^2 4x^3 - 3x^2 + C \, dx = 18$

8. Wow I can't believe I didn't see that.
So through intense calculations, I end up with $C=19$ if I'm correct. So now I plug in the x values 1 and 2 and subtract and what not, right?