# Derivative of a function

• Mar 16th 2009, 05:25 PM
calc_student09
Derivative of a function
A function f is defined for all real numbers and has he following three properties.
1) f(1)=5
2) f(3)=21
3)for all real values of a and b, f(a+b)-f(a)=kab+2b^2

a)use a=1 and b=2 to find the value of k
f(1+2)-f(1)=k(1)(2)+2(2)^2
f(3)-f(1)=2k+8
21-5=2k+8
8=2k
k=4
Is that right?

B)find f'(3)
c) find f'(x) and f(x) for all real x
I need help figuring out how to do the other 2 parts.
• Mar 16th 2009, 06:24 PM
ibrox
If I am interpreting this correctly:

Part A: Not sure, forget how to grab a 3rd point..and construct the equation from there

Parts B and C: These are easy once you get the equation of a function (i.e. f(x) ). Once you have that, you can power rule to get the derivative and sub 3 into the derivative of the function to get your answer.

Ibrox
• Mar 16th 2009, 06:55 PM
mr fantastic
Quote:

Originally Posted by calc_student09
A function f is defined for all real numbers and has he following three properties.
1) f(1)=5
2) f(3)=21
3)for all real values of a and b, f(a+b)-f(a)=kab+2b^2

a)use a=1 and b=2 to find the value of k
f(1+2)-f(1)=k(1)(2)+2(2)^2
f(3)-f(1)=2k+8
21-5=2k+8
8=2k
k=4
Is that right?

Mr F says: Yes.

B)find f'(3)
c) find f'(x) and f(x) for all real x
I need help figuring out how to do the other 2 parts.

Let $\displaystyle a = b = x$: $\displaystyle f(2x) - f(x) = 6x^2 \Rightarrow f(x) = 2x^2 + c$ where c is a constant.
• Mar 17th 2009, 06:01 PM
OnMyWayToBeAMathProffesor
I also got k=4. But how would you do $\displaystyle f'(3)$? and i am sorry 'mr fantastic', i did not understand your process. could you please explain?
• Mar 17th 2009, 06:42 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
Let $\displaystyle a = b = x$: $\displaystyle f(2x) - f(x) = 6x^2 \Rightarrow f(x) = 2x^2 + c$ where c is a constant.

Quote:

Originally Posted by OnMyWayToBeAMathProffesor
I also got k=4. But how would you do $\displaystyle f'(3)$? and i am sorry 'mr fantastic', i did not understand your process. could you please explain?

Given $\displaystyle f(2x) - f(x) = 6x^2$, it's not unreasonable to assume that $\displaystyle f(x)$ has the form $\displaystyle \alpha x^2$. Substitute this into $\displaystyle f(2x) - f(x) = 6x^2$ to find that $\displaystyle \alpha = 2$.

Once you have the rule it's simple to get the value of the derivative.
• Mar 18th 2009, 05:52 PM
OnMyWayToBeAMathProffesor
i'm sorry but i fail to see how $\displaystyle f(2x) - f(x) = 6x^2$ is given. $\displaystyle a$ and $\displaystyle b$ are two different integers and i did not think that you could just combine the 2. and where does $\displaystyle 6x^2$ come in? I am sorry but I think i am very lost.
• Mar 18th 2009, 06:29 PM
mr fantastic
Quote:

Originally Posted by OnMyWayToBeAMathProffesor
i'm sorry but i fail to see how $\displaystyle f(2x) - f(x) = 6x^2$ is given. $\displaystyle a$ and $\displaystyle b$ are two different integers and i did not think that you could just combine the 2. and where does $\displaystyle 6x^2$ come in? I am sorry but I think i am very lost.

You can let a and b be anything you want. They are merely pronumerals.

Quote:

Originally Posted by mr fantastic
Let $\displaystyle a = b = x$: $\displaystyle f(2x) - f(x) = 6x^2 \Rightarrow f(x) = 2x^2 + c$ where c is a constant.

f(a + b) = f(x + x) = f(2x).

f(a) = f(x)

$\displaystyle 4ab + 2b^2 = 4x^2 + 2x^2 = 6x^2$.

And the result I posted follows.