Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter P - Problem Solving - Page 39: 3

Answer

See image for the graphs.

Work Step by Step

$ H(x) = \left\{ \begin{array}{lr} 1 & : x \geq 0\\ 0 & : x < 0 \end{array} \right.$ (a) $ H(x)-2 = \left\{ \begin{array}{lr} -1 & : x \geq 0\\ -2 & : x < 0 \end{array} \right.$ This is found by subtracting 2 from each of the outputs of the Heaviside function. $$1-2=-1$$ $$0-2=-2$$ (b) $ H(x-2) = \left\{ \begin{array}{lr} 1 & : x \geq 2\\ 0 & : x < 2 \end{array} \right.$ This is found by substituting $(x-2)$ in for $x$ in the Heaviside function and then solving for $x$. $$x-2\geq0 \text{ thus } x\geq 2 $$ $$x-2<0 \text{ thus } x< 2$$ (c) $ -H(x) = \left\{ \begin{array}{r} -1 & : x \geq 0\\ 0 & : x < 0 \end{array} \right.$ This is found by multiplying each of the outputs of the Heaviside function by -1. $$-1(1)=-1$$ $$-1(0)=0$$ (d) $ H(-x) = \left\{ \begin{array}{lr} 1 & : x \leq 0\\ 0 & : x > 0 \end{array} \right.$ This is found by substituting $-x$ in for $x$ in the Heaviside function and then solving for $x$. $$-x\geq0 \text{ thus } x\leq 0 $$ $$-x<0 \text{ thus } x>0$$ (e) $\frac{1}{2} H(x) = \left\{ \begin{array}{lr} \frac{1}{2} & : x \geq 0\\ 0 & : x < 0 \end{array} \right.$ This is found by multiplying each of the outputs of the Heaviside function by $\frac{1}{2}$. $$\frac{1}{2}(1)=\frac{1}{2}$$ $$\frac{1}{2}(0)=0$$ (f) $ -H(x-2)+2 = \left\{ \begin{array}{lr} 1 & : x \geq 2\\ 2 & : x < 2 \end{array} \right.$ This one has two parts. First, substitute $(x-2)$ in for $x$ in the Heaviside function and solve for $x$. $$x-2\geq0 \text{ thus } x\geq 2 $$ $$x-2<0 \text{ thus } x< 2$$ Second, multiply each of the outputs of the Heaviside function by $-1$ then add 2. $$-1(1)+2=1$$ $$-1(0)+2=2$$
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