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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 3: Q. 3.98 (page 323)

For the function $h (x) = 2 x + 1$ , evaluate the function.
$h (k^{2})$
$h (x + 1)$
$h (x) + h (1)$

Short Answer

Expert verified

Part (a) $h (k^{2}) = 2 k^{2} + 1$

Part (b) $h (x + 1) = 2 x + 3$

Part (c) $h (x) + h (1) = 2 x + 4$

Step by step solution

01

Part (a) Step 1.

Substitute $x = k^{2}$ in localid="1644749461383" $h (x) = 2 x + 1$ ,

$\begin{array}{rcl} h (k^{2}) & = & 2 (k^{2}) + 1 \\ = & 2 k^{2} + 1 \end{array}$

02

Part (b) Step 1.

Substitute $x + 1$ for $x$ in localid="1644749540514" $h (x) = 2 x + 1$ ,

$\begin{array}{rcl} h (x + 1) & = & 2 (x + 1) + 1 \\ = & 2 x + 2 + 1 \\ = & 2 x + 3 \end{array}$

03

Part (c) Step 1.

$\begin{array}{rcl} h (x) + h (1) & = & (2 x + 1) + (2 (1) + 1) \\ = & 2 x + 1 + (2 + 1) \\ = & 2 x + 1 + 3 \\ = & 2 x + 4 \end{array}$

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