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Exercises Functions

Example 8 from Algebra and Trigonometry, 3.1 Functions and Function Notation

Solving Functions

Given the function $h (p) = p^{2} + 2 p,$ solve for $h (p) = 3.$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
$\begin{array}{l} h (p) = 3 \\ p^{2} + 2 p = 3 & Substitute the original function h (p) = p^{2} + 2 p . \\ p^{2} + 2 p - 3 = 0 & Subtract 3 from each side . \\ (p+3)(p-1)=0 & Factor . \end{array}$

If $(p + 3) (p - 1) = 0,$ either $(p + 3) = 0$ or $(p - 1) = 0$ (or both of them equal 0). We will set each factor equal to 0 and solve for $p$ in each case.

$\begin{array}{l} (p + 3) = 0, & p = - 3 \\ (p - 1) = 0, & p = 1 \end{array}$
This gives us two solutions. The output $h (p) = 3$ when the input is either $p = 1$ or $p = - 3.$ We can also verify by graphing as in Figure 3. The graph verifies that $h (1) = h (- 3) = 3$ and $h (4) = 24.$

Figure 3

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Given the function $g (m) = \sqrt{m - 4},$ solve $g (m) = 2.$

How did it go?

PreviousEvaluating Functions Next Finding an Equation of a Function

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 3.1, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.