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Section 1.1 : Review : Functions
3. Perform the indicated function evaluations for \(h\left( z \right) = \sqrt {1 - {z^2}} \).
- \(h\left( 0 \right) \)
- \(h\left( { - \frac{1}{2}} \right)\)
- \(h\left( {\frac{1}{2}} \right) \)
- \(h\left( {9z} \right) \)
- \(h\left( {{z^2} - 2z} \right) \)
- \(h\left( {z + k} \right) \)
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a \(h\left( 0 \right) \) Show Solution
\[h\left( 0 \right) = \sqrt {1 - {0^2}} = \sqrt 1 = 1\]
b \(h\left( { - \frac{1}{2}} \right)\) Show Solution
\[h\left( { - \frac{1}{2}} \right) = \sqrt {1 - {{\left( { - \frac{1}{2}} \right)}^2}} = \sqrt {\frac{3}{4}} = \frac{{\sqrt 3 }}{2}\]
c \(h\left( {\frac{1}{2}} \right) \) Show Solution
\[h\left( {\frac{1}{2}} \right) = \sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} = \sqrt {\frac{3}{4}} = \frac{{\sqrt 3 }}{2}\]
Hint : Don’t let the fact that there are new variables here instead of numbers get you confused. This works exactly the same way as the first three it will just have a little more algebra involved.
\[h\left( {9z} \right) = \sqrt {1 - {{\left( {9z} \right)}^2}} = \sqrt {1 - 81{z^2}} \]
Hint : Don’t let the fact that there are now variables here instead of numbers get you confused. This works exactly the same way as the first three it will just have a little more algebra involved.
\[h\left( {{z^2} - 2z} \right) = \sqrt {1 - {{\left( {{z^2} - 2z} \right)}^2}} = \sqrt {1 - \left( {{z^4} - 4{z^3} + 4{z^2}} \right)} = \sqrt {1 - 4{z^2} + 4{z^3} - {z^4}} \]
Hint : Don’t let the fact that there are now variables here instead of numbers get you confused. Also, don’t get excited about the fact that there is both a \(z\) and a \(k\) here. This works exactly the same way as the first three it will just have a little more algebra involved.
\[h\left( {z + k} \right) = \sqrt {1 - {{\left( {z + k} \right)}^2}} = \sqrt {1 - \left( {{z^2} + 2zk + {k^2}} \right)} = \sqrt {1 - {z^2} - 2zk - {k^2}} \]