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Suppose the function $f(x)$ is defined on the domain $\{x_1,x_2,x_3\}$, so that the graph of $y=f(x)$ consists of just three points. Suppose those three points form a triangle of area $32$.

The graph of $y = 2f(2x)$ also consists of just three points. What is the area of the triangle formed by those three points?

off-topic
Mar 27, 2019
edited by Guest  Mar 27, 2019
edited by Guest  Mar 27, 2019

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$$\text{Each of your new points is derived as }\\ (x^\prime_k,y^\prime_k) = \begin{pmatrix}\frac 1 2 &0 \\0 &2\end{pmatrix}\begin{pmatrix}x_k\\y_k\end{pmatrix}$$

$$\text{This results in an overall transformation matrix of }\\ \begin{pmatrix} \frac 1 2 &0 &0 &0 &0 &0\\ 0 &2 &0 &0 &0 &0 \\ 0 &0 &\frac 1 2 &0 &0 &0\\ 0 &0 &0 &2 &0 &0 \\ 0 &0 &0 &0 &\frac 1 2 &0 \\ 0 &0 &0 &0 &0 &2 \end{pmatrix}$$

$$\text{The determinant of this matrix is simply the product of the diagonal terms and equals }1\\ \text{Thus the transformation will leave the original area unchanged}$$

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Mar 27, 2019