Question

Suppose f is a real-valued differentiable function defined on $\left[ {1,\infty } \right]$ with $f(1) = 1$ . Moreover , suppose that f satisfies $f'(x) = \dfrac{1}{{{x^2} + {f^2}(x)}}$ , show that $f(x) < 1 + \dfrac{\pi }{4}\forall x \geqslant 1$ .

Answer 1

Hint:As it is given that $f'(x) = \dfrac{1}{{{x^2} + {{\left( {f(x)} \right)}^2}}} > 0\forall x \geqslant 1$ hence $f'(x) > 0$ $\forall x \geqslant 1$ then
$f(x)$ is an increasing function $\forall x \geqslant 1$ so we can write as $f'(x)$ as $f'(x) \leqslant \dfrac{1}{{1 + {x^2}}}{\text{ }}\forall {\text{ x }} \geqslant {\text{ 1}}$ . Now integrated $f'(x)$ from $1$ to some value $x$. As we have to prove that $f(x) < 1 + \dfrac{\pi }{4}\forall x \geqslant 1$ it mean that the maximum value of $f(x)$ is less than $1 + \dfrac{\pi }{4}$ solve according to that .

Complete step-by-step answer:
As we know that the square of any number is positive hence in the question , the denominator part of $f'(x)$ is positive
hence ,
$f'(x) = \dfrac{1}{{{x^2} + {{\left( {f(x)} \right)}^2}}} > 0\forall x \geqslant 1$
As we know the if $f'(x) > 0$ $\forall x \geqslant 1$ then
$ \Rightarrow $ $f(x)$ is an increasing function $\forall x \geqslant 1$
It is given that $f(1) = 1$ implies that $f(x) \geqslant 1{\text{ }}$ for all the value of ${\text{ }}x \geqslant 1$ because $f(x)$ is an increasing function $\forall x \geqslant 1$
As from the above the minimum value of $f(x)$ is $1$ hence we can write $f'(x)$ as
$f'(x) \leqslant \dfrac{1}{{1 + {x^2}}}{\text{ }}\forall {\text{ x }} \geqslant {\text{ 1}}$ we $f(x) = 1$ because it his minimum value and $f(x)$ is an increasing function $\forall x \geqslant 1$ so the value of $f'(x)$ decreasing as $f(x)$ is increasing.
So
$f'(x) \leqslant \dfrac{1}{{1 + {x^2}}}{\text{ }}\forall {\text{ x }} \geqslant {\text{ 1}}$
Now integrated $f'(x)$ from $1$ to some value $x$ where $x$ is variable
$\int\limits_1^x {f'(x)} \leqslant \int\limits_1^x {\dfrac{1}{{1 + {x^2}}}} {\text{ }}$
 Hence integration of $f'(x)$ is $f(x)$ and integration of $\dfrac{1}{{1 + {x^2}}}$ is ${\tan ^{ - 1}}x$
$f(x) - f(1) \leqslant {\tan ^{ - 1}}x - {\tan ^{ - 1}}1$
As we know that $f(1) = 1$ and ${\tan ^{ - 1}}x = \dfrac{\pi }{4}$ so by putting ,
$f(x) - 1 \leqslant {\tan ^{ - 1}}x - \dfrac{\pi }{4}$
$f(x) \leqslant {\tan ^{ - 1}}x - \dfrac{\pi }{4} + 1$
As we have to prove that $f(x) < 1 + \dfrac{\pi }{4}\forall x \geqslant 1$ it mean that the maximum value of $f(x)$ is less than $1 + \dfrac{\pi }{4}$ ,
Hence in general we have to find the maximum value of $f(x)$ so for the maximum value ${\tan ^{ - 1}}x = \dfrac{\pi }{2}$ it is his maximum value or we will write it as ${\tan ^{ - 1}}x < \dfrac{\pi }{2}$
So,
$f(x) < \dfrac{\pi }{2} - \dfrac{\pi }{4} + 1$
$f(x) < \dfrac{\pi }{4} + 1{\text{ }}\forall {\text{ x > 1}}$.
Hence proved.

Note:For continuous function $f(x)$ , if $f'(x)$ = \[0\] or $f'''(x) = 0$ does not exist at points $f'(x)$ where exists and if $f''(x)$ changes sign when passing through $x = x_0$ then $x_0$ is called the point of inflection.
(a) If $f''(x) < 0,x \in \left( {a,b} \right)$ then the curve $y = f(x)$ in concave downward
(b) if $f''(x) > 0,x \in \left( {a,b} \right)$ then the curve $y = f(x)$ is concave upwards in $\left( {a,b} \right)$

As if $f'(x) < 0$ then $f(x)$ is a decreasing function.