Exemplo 3 de Algebra and Trigonometry, 3.5 Transformation of Functions

Adding a Constant to an Input

Problema e solução em inglês, como no livro original.

Returning to our building airflow example from Figure 3, suppose that in autumn the facilities manager decides that the original venting plan starts too late, and wants to begin the entire venting program 2 hours earlier. Sketch a graph of the new function.

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Solução
We can set $V (t)$ to be the original program and $F (t)$ to be the revised program.

$\begin{matrix} V (t) & = & the original venting plan \\ F (t) & = & starting 2 hrs sooner \end{matrix}$
In the new graph, at each time, the airflow is the same as the original function $V$ was 2 hours later. For example, in the original function $V,$ the airflow starts to change at 8 a.m., whereas for the function $F,$ the airflow starts to change at 6 a.m. The comparable function values are $V (8) = F (6) .$ See Figure 6. Notice also that the vents first opened to $220 {ft}^{2}$ at 10 a.m. under the original plan, while under the new plan the vents reach $220 {ft}^{2}$ at
8 a.m., so $V (10) = F (8) .$
In both cases, we see that, because $F (t)$ starts 2 hours sooner, $h = - 2.$ That means that the same output values are reached when $F (t) = V (t - (- 2)) = V (t + 2) .$

Figure 6

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Solução

Solução