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Suppose $f(9)=2$.For each of the part below,find a point that must be on the graph of the given equation.

$a) \space y=f(x-3)+5 \\b)y=2f(x/4)\\ c)\space y=2f(3x-1)+7$

My attempt

Part $(a)$

I have that $(9,f(9))$ is a point on the original function $y=f(x)$ ,now the new function is $y=f(x-3)+5$ ,therefore I have $$(9,f(9)) \mapsto (12,f(9)+5) \mapsto (12,7)$$

Part $(b)$

The new function is now $y=2f(x/4)$ ,so I have $$(9,f(9)) \mapsto (36,f(9)) \mapsto (36,2f(9)) $$

Part $(c)$

The new function is $y=2f(3x-1)+7$,applying respectively the transformations: $y=f(3x),y=f(3x-1),y=2f(3x-1),y=2f(3x-1)+7$ ,I have $$ \left(9,f(9)\right) \mapsto \left(\cfrac{9}{3},f(9)\right) \mapsto \left(\cfrac{9}{3} +1,f(9)\right) \mapsto \left(\cfrac{9}{3}+1,2f(9)\right) \mapsto \left(\cfrac{9}{3} +1,2f(9)+7\right)$$

My book doesn't have solutions ,and I would like to be sure about my understanding of the previous material ...so if someone would check up my calculations I would be really grateful .

Thanks in advance

Mr. Y
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1 Answers1

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In part (b), the $2f(9)$ is in fact $2\times 2=4$.

In part (c), the $\frac{9}{3}+1$ should be $\frac{9+1}{3}$, and similarly as in part (b), you can calculate the $2f(9)+7$.

Edit:

If you do transformation for $f(3x-1)$, since this is on $x$, the order is reversed:

Step 1: $f(x)\rightarrow f(x-1)$

Step 2: $f(x-1)\rightarrow f(3x-1)$

Or you can treat it as $f(3(x-\frac{1}{3}))$. In this case, still using the reversed order:

Step 1: $f(x)\rightarrow f(3x)$

Step 2: $f(3x)\rightarrow f(3(x-\frac{1}{3}))$

KittyL
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  • Why should it be $\cfrac{9+1}{3}$ ?$f(3x-1)$ is $1$ to the right of $f(3x)$ so every point $(a,f(a)) \mapsto (a+1,f(a))$.What am I missing here ? – Mr. Y Jan 02 '16 at 20:22
  • @Mr.Y I am not sure about your transformation notation. Basically you want $3x-1=9$ so that $f(3x-1)=f(9)$, that's how you get $x=\frac{9+1}{3}$. – KittyL Jan 02 '16 at 21:28
  • Oh,alright.So I can't apply transformation linearly ,one after the other .As I have done ,I have considered in each case a different kind of function so that lead me to the wrong result,is that correct ? – Mr. Y Jan 03 '16 at 07:21
  • @Mr.Y On second thought, you can still do transformation. I added the transformation in the edit. – KittyL Jan 03 '16 at 10:24