Question

f(x + y) = f(x) + f(y) + xy^2 + x^2y and lim(x -->0) f(x)/x = 1, then find value of f' (3)

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$f(x + y) = f(x) + f(y) + x {y}^{2} + {x}^{2} y$

Putting x = 0 & y = 0 in both sides we get

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

$\implies \: f(0) = 0$

Now it is given that

$\displaystyle \lim_{x \to 0} \frac{f(x)}{x} = 1$

So

$f'(0)$

$= \displaystyle \lim_{x \to 0} \frac{f(x) - f(0)}{x}$

$= \displaystyle \lim_{x \to 0} \frac{f(x) }{x}$

$= 1$

Now

$f(x + y) = f(x) + f(y) + x {y}^{2} + {x}^{2} y$

Putting y = h

$f(x + h) = f(x) + f(h) + x {h}^{2} + {x}^{2} h$

So

$f'(x)$

$= \displaystyle \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$= \displaystyle \lim_{h \to 0} \frac{ f(x) + f(h) + x {h}^{2} + {x}^{2} h- f(x)}{h}$

$= \displaystyle \lim_{h \to 0} \frac{ f(h) + x {h}^{2} + {x}^{2} h}{h}$

$= \displaystyle \lim_{h \to 0} \frac{ f(h) }{h} + \displaystyle \lim_{h \to 0} \frac{ x {h}^{2} + {x}^{2} h}{h}$

$= 1 + \displaystyle \lim_{h \to 0} ( x {h} + {x}^{2} )$

$= 1 + {x}^{2}$

Hence

$f'(3) = 1 + {(3)}^{2} = 1 + 9 = 10$