Question

Let $f:R \to R$ is a function satisfying the condition $f(x + {y^3}) = f(x) + {[f(y)]^3}$ for all $x,y \in R$. If $f'(0) \geqslant 0$ then find $f(10)$.

Answer 1

Substitute the values of $x,y$ as zero in the given equation to find the value of $f(0)$. From this value, calculate the value of $f'(0)$ using the formula of limits. Finally, find the value of $f(x)$ from these values and substitute the value of $x$ as $10$ to find $f(10)$.

Complete step by step answer:
It is given to us that $f(x + {y^3}) = f(x) + {[f(x)]^3}$ and $f'(0) \geqslant 0$
In order to find the value of $f(0)$ let us substitute $x = 0,y = 0$ in the given equation.
The equation now becomes $f(0 + 0) = f(0) + {[f(0)]^3}$ which is $f(0) = f(0) + f(0)$
Therefore, the value of $f(0)$ is zero. Now, let us calculate the value of $f'(0)$
$f'(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(h)}}{h}$ since the value of $f(0)$ is zero.

Let us assume that this equation is $I$ and hence $I = f'(0)$
We can also write this value as
\Rightarrow I = f'(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + {{\left( {{h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}}} \right)}^3}) - f(0)}}{{{{({h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}})}^3}}}

Since the value of $f(0)$ is zero, the above expression becomes
\Rightarrow I = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + {{\left( {{h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}}} \right)}^3})}}{{{{({h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}})}^3}}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{[f({h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}})]}^3}}}{{{{({h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}})}^3}}}
This expression can be written as
\Rightarrow I = \mathop {\lim }\limits_{h \to 0} {\left( {\dfrac{{f\left( {{h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}}} \right)}}{{\left( {{h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}}} \right)}}} \right)^3} = {I^3}
For $I = {I^3}$ the possible values of $I$ are $- 1,1,0$ but it is already given to us that $f'(0) \geqslant 0$ so the possible values become $0,1$
In the same way we can calculate the value of $f'(x)$
$\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$ and we write this value as
\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + {{\left( {{h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}}} \right)}^3}) - f(x)}}{{{{\left( {{h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}}} \right)}^3}}}

From the given equation,
\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x) + {{[f\left( {{h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}}} \right)]}^3} - f(x)}}{{{{\left( {{h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}}} \right)}^3}}}
By solving, we get
\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} {\left( {\dfrac{{f({h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}})}}{{{h^{{\raise0.7ex\hbox{1} \\!\mathord{\left/ {\vphantom {1 3}}\right.} \\!\lower0.7ex\hbox{3}}}}}}} \right)^3} = {I^3} = {\left( {f'(0)} \right)^3}

We already know the possible values of ${I^3}$ so now the possible values of $f'(x)$ are $0,1$.
If $f'(x) = 0$, by integrating we get $f(x) = c$ where c is the constant.
If $f'(x) = 1$, by integrating we get $f(x) = x + c$
Since $f(0) = 0$ the value of c would be zero.
Now, if we substitute $x = 10$, $f(10) = 0$ or $f(10) = 10$

Therefore the value of $f(10)$ is $0$ or $10$.

Note: It should be noted that since $f'(0) \geqslant 0$, $f'(0)$ could have multiple values so one should not make a mistake by considering only one value. Similarly when $f'(x)$ is integrated, one should not forget to take the constant c into consideration.