Some function

Algebra Level 4

$f\left( 1+\frac { \sqrt { 2 } }{ x } \right) =\frac { \left( 1+\sqrt { 2017 } \right) { x }^{ 2 }+2\sqrt { 2 } x+2 }{ { x }^{ 2 } } ,\quad x\neq 0$

Let function $f$ be defined as above. Find the value of $f\left( \sqrt { 2018-\sqrt { 2017 } } \right)$

The answer is 2018.00.

1 solution

Chew-Seong Cheong
Jun 26, 2017

$\begin{aligned} f \left(1 + \frac {\sqrt 2}x\right) & = \frac{\left(1+\sqrt{2017}\right)x^2+2\sqrt 2 x + 2}{x^2} \\ & = 1+\sqrt{2017} + \frac {2\sqrt 2}x + \frac 2{x^2} \\ & = \sqrt{2017} + \left(1 + \frac {\sqrt 2}x\right)^2 \\ \implies f \left({\color{#3D99F6}\sqrt{2018-\sqrt{2017}}} \right) & = \sqrt{2017} + \left({\color{#3D99F6}\sqrt{2018-\sqrt{2017}}} \right)^2 \\ & = \boxed{2018} \end{aligned}$

Note that when $1 + \dfrac {\sqrt 2}x = \sqrt{2018-\sqrt{2017}}$ $\implies x \approx 0.032570957$ , substituting it in the equation we get 2018 as answer.

For completeness, we should ensure that there is a corresponding value of $x$ which gives us $1 + \frac { \sqrt{2} } { x} = \sqrt{ 2018 - \sqrt{2017} }$ .

For example, we cannot determine what $f(1)$ would be equal to.

Calvin Lin Staff - 3 years, 11 months ago

Thanks. I have done that.

Chew-Seong Cheong - 3 years, 11 months ago

f(1)=sqrt(2017)+1

Chukwuebuka Daniel - 1 year ago

What value of $x$ allows you to draw that conclusion?

Are you assuming that the function is continuous?

Calvin Lin Staff - 1 year ago

Some function

The answer is 2018.00.

1 solution

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