Using Properties to determine derivative and original equation

**ment2byours** · March 1st 2009, 09:30 AM

A function f is defined for all real numbers and has the following properties.

(i) f(1)=5
(ii) f(3)=21
(iii) for all real values of a and b, f(a+b) - f(a)= kab+2b^2 where k is a fixed real number independent of a and b

a) Use a=1 and b=2 to find k.
b) Find f'(3)
c) Find f'(x) and f(x) for all real x

So I found out that k=-4, but I'm stuck on how to find the original equation and thus the derivative by the given info.

**redsoxfan325** · March 1st 2009, 04:48 PM

Originally Posted by ment2byours

A function f is defined for all real numbers and has the following properties.

(i) f(1)=5
(ii) f(3)=21
(iii) for all real values of a and b, f(a+b) - f(a)= kab+2b^2 where k is a fixed real number independent of a and b

a) Use a=1 and b=2 to find k.
b) Find f'(3)
c) Find f'(x) and f(x) for all real x

So I found out that k=-4, but I'm stuck on how to find the original equation and thus the derivative by the given info.

First of all, for (a), I think $k=4$ , not $-4$ . Here's why: $21-5=k*1*2+2*2^2$ . Thus $16=2k+8$ and $k=4$ .

Use the definition of the derivative (using a and b for x and h): $\lim_{b\to 0} \frac{f(a+b)-f(a)}{b}$

You know $f(a+b)-f(a)=4ab+2b^2$ so the above limit equals $\lim_{b\to 0} \frac{4ab+2b^2}{b}=4a$ .

Thus $f'(a) = 4a$ . Thus $f'(3)=12$ .

**ment2byours** · March 1st 2009, 04:53 PM

Oh thanks!! I should have recognized that it was similar for the definition for a derivative. Sorry about getting k wrong, I tend to keep messing up my pos/neg signs. So the deriv of the original equation is f'(x)=4x?

**arpitagarwal82** · March 1st 2009, 04:55 PM

Continuing with above solution

Now we have f'(x) = 4x

So $f(x) = 2x^2 + c$ where c is constant of integration

using f(1) = 5
5 = 2 + c

c = 3

so function is $f(x) = 2x^2 + 3$