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# Suppose $f(x)$ is a rational function such that $3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$ for all $x \neq 0$. Find $f(-2)$.

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Suppose f(x) is a rational function such that $$3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$$ for all x =\=0. Find f(-2).

May 22, 2018

### 3+0 Answers

#1
+100006
+1

Suppose f(x) is a rational function such that

$$3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$$

for all $$x\ne0$$. Find f(-2).

We have

$$3 f \left( \frac{1}{-2} \right) + \frac{2f(-2)}{-2} = (-2)^2\\ 3 f \left( \frac{1}{-2} \right) - f(-2) =4\qquad \qquad(1)\\ and\\ 3 f \left( \frac{1}{-1/2} \right) + \frac{2f(-1/2)}{-1/2} = (-1/2)^2\\ 3 f \left( -2 \right) -4f(\frac{-1}{2}) =\frac{1}{4}\\ -4f(\frac{-1}{2}) +3 f \left( -2 \right)=\frac{1}{4}\qquad\qquad (2)\\--------------\\~\\ \quad 3 f \left( \frac{1}{-2} \right) - f(-2) =4\qquad \qquad(1)\\ \quad 12f \left( \frac{1}{-2} \right) - 4f(-2) =16\qquad \qquad(1b)\\ -4f(\frac{-1}{2}) +3 f \left( -2 \right)=\frac{1}{4}\qquad\qquad (2)\\ -12f(\frac{-1}{2}) +9 f \left( -2 \right)=\frac{3}{4}\qquad\qquad (2b)\\~\\ (1b)+(2b)\\ 5f(-2)=16\frac{3}{4}\\ f(-2)=3\frac{7}{20}\\$$

The method is correct but you need to check for careless errors.

May 22, 2018
#2
+21977
+1

Suppose

$$f(x)$$

is a rational function such that  $$3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$$

for all $$x \neq 0$$. Find $$f(-2)$$.

$$\begin{array}{|lrclcl|} \hline & 3f(\frac1x) + \frac2x f(x) &=& x^2 \\ & 3f(\frac1x) &=& x^2 - \frac2x f(x) \\ & f(\frac1x) &=& \frac{x^2}{3} - \frac{2}{3x} f(x) \qquad (1) \\ \\ \text{Set }x=\frac1x: & 3f(x) + 2x f(\frac1x) &=& \frac{1}{x^2} \\ & 2x f(\frac1x) &=& \frac{1}{x^2}-3f(x) \\ & f(\frac1x) &=& \frac{1}{2x^3}-\frac{3}{2x}f(x) \qquad (2) \\\\ \hline (1) = (2): & \frac{x^2}{3} - \frac{2}{3x} f(x) &=& \frac{1}{2x^3}-\frac{3}{2x}f(x) \\ & \frac{3}{2x}f(x) - \frac{2}{3x} f(x) &=& \frac{1}{2x^3} -\frac{x^2}{3} \\ & f(x)\left( \frac{3}{2x} - \frac{2}{3x} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{9x-4x}{6x^2} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{5x}{6x^2} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{5x}{1} \right) &=& \frac{3-2x^5}{x} \\\\ & \mathbf{f(x)} & \mathbf{=} & \mathbf{\dfrac{3-2x^5}{5x^2}} \\ \hline \end{array}$$

The rational function is  $$f(x) = \dfrac{3-2x^5}{5x^2}$$

$$f(-2)=\ ?$$

$$\begin{array}{|rclcl|} \hline f(-2) &=& \frac{3-2(-2)^5}{5(-2)^2} \\ &=& \frac{3+2^6}{5\cdot 4} \\ &=& \frac{3+64}{20} \\ &=& \frac{67}{20} \\ \mathbf{f(-2)} & \mathbf{=} & \mathbf{3.35} \\ \hline \end{array}$$

$$\text{ f(-2) is 3.35 }$$

graph:

May 22, 2018
edited by heureka  May 22, 2018
#3
+100006
+1

That is interesing Heureka, I had not realised that there was enough infromation to draw the graph. :)

Melody  May 22, 2018