$g(x)=f(6x)+1$. Range is given for $f(x)$. Find Range of $g(x)$.

Question

Suppose a function $f(x)$ has domain $(-\infty,\infty)$ and range $[-11,3]$. If we define a new function $g(x)$ by $$g(x) = f(6x)+1,$$ then what is the range of $g(x)$? Express your answer in interval notation.

First think I thought of doing was multiplying $-11$ by $6$ and adding $1$ to get $-65$. Then I did the same thing for $3$ to get $19$. I thought the answer was $[-65,19]$. However, this was marked as wrong...Where am I going wrong?

Answer 1

Range of $f: [-11,3]$,

Domain of $f: (-\infty,+\infty).$

Means: For every $x \in D=(-\infty,\infty)$ :

$-10 \le f(x) \le 3.$

For every $6x \in D = (-\infty,\infty)$ :

$-10 \le f(6x) \le 3.$

Range of $f(x)$ = Range of $f(6x)$.

Range of $g(x)$ = Range of $(f(6x)+1)$

for $x \in D= (-\infty, \infty)$:

Every image point of $f(6x)$ is shifted up by $+1$:

Hence: $[-10,4]$.

Answer 2

The function $f(x)$ can attain values on $[-11,3]$.

Hence the function $f(6x)$ can attain values on $[-11,3]$.

After that, we perform a vertical shift, the functon $f(6x)+1$ can attain values on $[-10, 4]$.

Some transformation scale things horizontally (such as $f(6x)$) and some functions scale things affect things vertically.

Answer 3

The domain is unchanged, while the range becomes $[-10, 4]$.

This is because your original domain is $-\infty < x < \infty$. When you plug in $6x$, it must satisfy $-\infty < 6x < \infty$ because we're plugging $6x$ in place of $x$ which is still $-\infty < x < \infty$ thus the domain is unchanged.

As for the range, $g(x)$ is adding $1$ to the output of $f(6x)$. Since $f(6x)$ will only give you values in $[-11, 3]$, because that is the range of $f(x)$, $g(x)$ will only ever give you values in $[-11+1, 3+1]$.