Question

$f(x)$ is a polynomial function. Two conditions are given.
$\boxed{\begin{aligned}&(1)\ f(a+b)=f(a)+f(b)+3xy\\&(2)\ f'(0)=1\end{aligned}}$
Find $f(x)$.

Answer 1

$\Large\text{\underline{\underline{Question}}}$

We have two conditions: -

$\text{ \cdots\longrightarrow\boxed{f(a+b)=f(a)+f(b)+3ab.} \cdots (1)}$

$\text{ \cdots\longrightarrow\boxed{f'(0)=1.} \cdots (2)}$

Find such a polynomial function $f(x)$.

$\Large\text{\underline{\underline{Explanation}}}$

Main concept

Let's use the first principle of derivative -

$\text{ \cdots\longrightarrow\displaystyle\boxed{\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}. [The first principle of derivative] } }$

$\Huge\text{[1]}$

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}\displaystyle&f'(x)\\\\&=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\\\\&=\lim_{h\to0}\dfrac{f(x)+f(h)+3hx-f(x)}{h}\\\\&=\lim_{h\to0}\dfrac{f(h)+3hx}{h}\\\\&=3x+\lim_{h\to0}\dfrac{f(h)}{h}. \end{aligned}} }$

$\Huge\text{[2]}$

Let's put $x=0$. It is given in (2), -

$\text{ \cdots\longrightarrow\boxed{\displaystyle&\lim_{h\to0}\dfrac{f(h)}{h}=1.} }$

$\Huge\text{[3]}$

Then, by properties of limits, $\text{ h\to 0}$ then $f(0)\to0$ as well. Since $f(x)$ is continuous everywhere, the limiting value is equal to the function value.

$\cdots\longrightarrow\boxed{f(0)=0.}$

$\Huge\text{[4]}$

We already found the derivative in [1]. By integration, -

$\cdots\longrightarrow\boxed{\begin{aligned}&f(x)\\\\\displaystyle&=\int f'(x) dx\\\\&=\int(3x+1)dx\\\\&=\dfrac{3}{2}x^{2}+x+C.\end{aligned}}$

Since $f(0)=0$ in [3], -

$\cdots\longrightarrow\boxed{C=0.}$

Conclusion

Hence by [4], -

$\cdots\longrightarrow\boxed{f(x)=\dfrac{3}{2}x^{2}+x.}$

Answer 2

is a polynomial function. Two conditions are given.

Find .