Write the formula for $f(x)$, if the graph of $f$ can be obtained from the graph of $y = g(x)$ by shrink horizontally by a factor of $5$ then shift left $3$ units The equation should be $f(x) = g(5(x+3))$ or $g\left(\frac{1}{5}(x+3)\right)$? I prefer the second answer but my teacher said the correct is the first one? Can anyone explain for me why it is $5$ instead of $\frac{1}{5}$ while we are dealing with horizontal shrinking? Thanks a lot

I found this counterintuitive when I was first learning algebra too.

Think about it like this: $f(5x)$ gives you $f(0)$ at $x=0$, then $f(5)$ at $x=1$, then $f(10)$ at $x=2$. Varying the input parameter from $0$ to $2$ made the function go all the way from $f(0)$ to $f(10)$. So the section of the graph of $f(x)$ that used to have width 10 will have only width 2 in the graph of $f(5x)$.

If that still doesn't click, I would just suggest drawing out a bunch of explicit examples for different functions $f$.

Intuitively, a function that's shrunk covers its original range values on a shorter interval. With $5x$ instead of $x$, consider the original function on the interval $[0,1]$, you get $5\cdot(1/5) =1$, so that the function covered all its original values on $[0,1]$ by the time you get to $x=1/5$, i.e on the interval $[0,1/5]$. In general then it covers its range 5 times faster.

Another easy way is to consider $cx$ for $c$ getting really large. Then for small $x$, you've already covered a huge portion of the function's range.

To shrink a function means to make the graph of the function seems narrower.

For example, consider the function $$f(x)=x^2$$ If you want to make the function shrink horizontally by a factor of 2 you would want the function $$f(2x) = (2x)^2 = 4x^2$$ On the other hand, you would argue that $$f\left(\frac{1}{2}x\right) = \left(\frac{1}{2}x\right)^2 = \frac{1}{4}x^2$$ is correct.

If you graph the functions, you would get

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Obviously, the function $f(2x) = (2x)^2 = 4x^2$ seems narrower. Similarly, your question is asking you to shrink the function by a factor of five, so it should be $f(5x)$ instead of $f\left(\frac{1}{5}x\right)$.

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