Interesting Mathematical Problems

Values of a rational function at irrational arguments

For the function f(x) = x²/(x−1) three values are calculated: at 1−√3, at 2−√3 and at the tangent of π/6. Each time the irrationality has to be removed from the denominator.

Finding a value of a function means substituting a number for x - simple, until that number is irrational. Then a second step appears: removing the root from the denominator by multiplying by the conjugate. Three examples show the same pattern in increasing difficulty.

The function is given

f(x) = \frac{x^{2}}{x - 1}

x is the variable - it changes its value.

Calculate f(1−√3), f(2−√3) and f(tan π/6).

1) The value at 1 − √3

f\left(1 - \sqrt{3}\right) = \frac{\left(1 - \sqrt{3}\right)^{2}}{1 - \sqrt{3} - 1}

= \frac{1 - 2\sqrt{3} + 3}{-\sqrt{3}} = \frac{4 - 2\sqrt{3}}{-\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}

= \frac{4\sqrt{3} - 6}{-3} = \frac{6 - 4\sqrt{3}}{3}

(a - b)^{2} = a^{2} - 2ab + b^{2}

2\sqrt{3} \cdot \sqrt{3} = 2 \cdot 3 = 6

2) The value at 2 − √3

f\left(2 - \sqrt{3}\right) = \frac{\left(2 - \sqrt{3}\right)^{2}}{2 - \sqrt{3} - 1}

= \frac{4 - 4\sqrt{3} + 3}{1 - \sqrt{3}} = \frac{7 - 4\sqrt{3}}{1 - \sqrt{3}} \cdot \frac{1 + \sqrt{3}}{1 + \sqrt{3}}

= \frac{7 + 7\sqrt{3} - 4\sqrt{3} - 12}{1 - 3} = \frac{-5 + 3\sqrt{3}}{-2} = \frac{5 - 3\sqrt{3}}{2}

(a - b)^{2} = a^{2} - 2ab + b^{2}

(a - b)(a + b) = a^{2} - b^{2}

4\sqrt{3} \cdot \sqrt{3} = 4 \cdot 3 = 12

3) The value at the tangent of π/6

f\left(\operatorname{tg} \tfrac{\pi}{6}\right) = \frac{\left(\operatorname{tg} \tfrac{\pi}{6}\right)^{2}}{\operatorname{tg} \tfrac{\pi}{6} - 1}

= \frac{\left(\tfrac{\sqrt{3}}{3}\right)^{2}}{\tfrac{\sqrt{3}}{3} - 1} = \frac{\tfrac{1}{3}}{\tfrac{\sqrt{3} - 3}{3}}

= \tfrac{1}{3} : \frac{\sqrt{3} - 3}{3} = \tfrac{1}{3} \cdot \frac{3}{\sqrt{3} - 3}

= \frac{1}{\sqrt{3} - 3} \cdot \frac{\sqrt{3} + 3}{\sqrt{3} + 3} = \frac{\sqrt{3} + 3}{3 - 9}

= \frac{3 + \sqrt{3}}{-6} = -\tfrac{1}{2} - \frac{\sqrt{3}}{6}

\pi = 180^{\circ}

\tfrac{\pi}{6} = \tfrac{180^{\circ}}{6} = 30^{\circ}

\operatorname{tg} 30^{\circ} = \tfrac{\sqrt{3}}{3}

\left(\tfrac{\sqrt{3}}{3}\right)^{2} = \tfrac{3}{9} = \tfrac{1}{3}

\tfrac{\sqrt{3}}{3} - 1 = \tfrac{\sqrt{3}}{3} - \tfrac{3}{3} = \tfrac{\sqrt{3} - 3}{3}

(a - b)(a + b) = a^{2} - b^{2}

Frequently asked questions

How do you remove an irrational number from a denominator?

Multiply the numerator and denominator by the conjugate of the denominator, that is the same expression with the sign reversed. The product of a sum and a difference gives a difference of squares, in which the root disappears.

What is the tangent of 30 degrees?

The square root of three divided by three. The angle π/6 in radians is exactly 30 degrees.

What is the variable in the notation of a function?

It is the slot into which successive numbers are substituted - hence the name: a variable changes its value. In f(x) = x²/(x−1) the variable is x and it appears in two places at once.

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A compound fraction with negative powers and logarithms

Determining the domain of a function