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For each point $(a,b)$ on the graph $y = f(x)$ the point $(3a-1,\frac{b}{2})$ is plotted forming the graph of another function $y = g(x)$. The problem asks us to find the function $g(x)$ in terms of $f(x)$.

Also, we need to describe the transformation.


Since $a$ goes to $3a-1$, the inverse of the function will be $x \to \frac{x+1}{3}$.

Hence the required function is $$g(x) = \frac{1}{2}f\left( \frac{x+1}{3}\right).$$

Note that $$g(3a-1) = \frac{1}{2}f\left( \frac{3a-1+1}{3}\right) = \frac{1}{2}f(a) = \frac{b}{2}.$$


Description of the transformation:

$\bullet$ Horizontal stretch by a factor of $3$.

$\bullet$ Vertical stretch by a factor of $\frac{1}{2}$.

$\bullet$ Horizontal translation by $-1$.


Is the solution correct?

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1 Answer 1

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Let $$X=3a-1\;\text{ and }\; g(X)=\frac b2$$

with

$$b=f(a) $$ or

$$2g(X)=f(\frac{X+1}{3})$$ So

$$g(X)=\frac 12f(\frac{X+1}{3})$$

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