MA005: Calculus I

Some compositions are relatively common and easy, and you should recognize the effect of the composition on the graphs of the functions.

Example 3: Fig. 6 shows the graph of $y = f(x)$.
Graph (a) $2 + f(x)$, (b) $3.f(x)$, and $(c) f(x – 1)$.

Solution: All of the new graphs are shown below in Fig. 7 .

(a) Adding 2 to all of the values of $f(x)$ rigidly shifts the graph of $f(x)$ 2 units upward.

(b) Multiplying all of the values of $f(x)$ by 3 leaves all of the roots of $f$ fixed: if $x$ is a root of $f$ then $f(x) = 0$ and $3f(x) = 3(0) = 0$ so $x$ is also a root of $3 . f(x)$. If $x$ is not a root of $f$, then the graph of $3f(x)$ looks like the graph of $f(x)$ stretched vertically by a factor of 3.

(c) The graph of $f(x–1)$ is the graph of $f(x)$ rigidly shifted 1 units to the right.

We could also get these results by examining the graph of $y = f(x)$, creating a table of values for $f(x)\$ and the new functions, and then graphing the new functions.

• the graph of $y = k + f(x)$ will be the graph of $y = f(x)$ rigidly shifted up by k units,
• the graph of $y = kf(x)$ will have the same roots as the graph of $f(x)$ and will be the graph of $y = f(x)$ vertically stretched by a factor of $k$,
• the graph of $y = f(x – k)$ will be the graph of $y = f(x)$ rigidly shifted right by $k$ units,
• the graph of $y = f(x + k)$ will be the graph of $y = f(x)$ rigidly shifted left by $k$ units.

Practice 5: Fig. 8 is the graph of $g(x)$.
Graph (a) $1+g(x)$, (b) $2g(x)$, (c) $g(x–1)$ and (d) $–3g(x)$.