A calculus problem by Harnakshvir Singh Dhillon

$\begin{aligned} f(x+y) & = f(x) + f(y) + xy & \small \color{#3D99F6} \text{Given; putting }x=y=0 \\ f(0) & = 2f(0) \\ \implies f(0) & = 0 & \small \color{#3D99F6} \text{Putting }y=h \\ f(x+h) & = f(x) + f(h) + xh & \small \color{#3D99F6} \text{Rearranging and} \\ \frac {f(x+h)-f(x)}h & = \frac {f(h)}h + x & \small \color{#3D99F6} \text{dividing both sides by }h \\ \lim_{h \to 0} \frac {f(x+h)-f(x)}h & = {\color{#3D99F6}\lim_{h \to 0} \frac {f(h)}h} + x & \small \color{#3D99F6} \text{Taking limit of }h \to 0 \\ f'(x) & = {\color{#3D99F6} 3} + x & \small \color{#3D99F6} \text{Given that }\lim_{h \to 0} \frac {f(h)}h = 3 \\ \implies f(x) & = 3x + \frac {x^2}2 + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ f(0) & = 0 & \small \color{#3D99F6} \implies C = 0 \\ \implies f(x) & = 3x + \frac {x^2}2 & \small \color{#3D99F6} \text{And } \\ f(x+y) & = 3x + \frac {x^2}2 + 3y + \frac {y^2}2 + xy & \small \color{#3D99F6} \implies f(x+y) = f(x) + f(y) + xy \text{ as given.} \\ \implies f(1) & = 3+\frac 12 = \boxed {\dfrac 72} \end{aligned}$

On putting x=y=0,

f(0)=f(0)+f(0)

=> f(0)=0

f ' (x) = lim(h-->inf) [{f(x+h)-f(x)}/ {(x+h)- x}]

and

f(x+h)= f(x)+f(h)+xh (from given identity)

therefore,

f ' (x)= lim(h-->inf) {f(h)+xh}/h
= lim(h-->inf) f(h)/h + lim(h-->inf) xh/h
= 3+x

On integrating,

f(x)= (x^2)/2 + 3x + c

but f(0)=0

==> f(x)= (x^2)/2+3x

==>f(1)=7/2

Replacing $x=y=0$ we get $f(0) = 2f(0) \implies f(0) = 0.$ Thus $\lim_{h \to 0} \frac{f(h)}{h} = \lim_{h \to 0} \frac{f(0 + h) - f(h)}{h} = f'(0)$ . So $f'(0) = 3$ .

Now differentiate the functional equation with respect to $x$ to get $f'(x+y) = f'(x) + y$ . Evaluate at $x=0$ to get $f'(y) = 3 + y$ . Integrate with respect to $y$ to get $f(y) = 3y + \frac{y^2}{2} + C$ . Because $f(0) = 0$ we know that $C = 0$ . The solution to the functional equation is thus $f(y) = 3y + \frac{y^2}{2}$ . And $f(1) = \frac{7}{2}$ .

At first differentiate the function wrt x keeping y constant. Then put x in places of y and 0 in place of x. From this we get f '(x)=f '(0)+x. From putting x=y=0 in the parental eqn we get f(0)=0 rewriting f(h)=f(h+0)-f(0), we get from the limit that f '(0)=3. Hence we get fx as=x+3 and integrating and comparing withf0 we get fx.