Paul's Online Notes
Home / Calculus I / Review / Functions
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 1-1 : Review : Functions

3. Perform the indicated function evaluations for $$h\left( z \right) = \sqrt {1 - {z^2}}$$.

1. $$h\left( 0 \right)$$
2. $$h\left( { - \frac{1}{2}} \right)$$
3. $$h\left( {\frac{1}{2}} \right)$$
1. $$h\left( {9z} \right)$$
2. $$h\left( {{z^2} - 2z} \right)$$
3. $$h\left( {z + k} \right)$$

Show All Solutions Hide All Solutions

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.
d $$h\left( {9z} \right)$$ Show Solution
$h\left( {9z} \right) = \sqrt {1 - {{\left( {9z} \right)}^2}} = \sqrt {1 - 81{z^2}}$

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.
e $$h\left( {{z^2} - 2z} \right)$$ Show Solution
$h\left( {{z^2} - 2z} \right) = \sqrt {1 - {{\left( {{z^2} - 2z} \right)}^2}} = \sqrt {1 - \left( {{z^4} - 4{z^3} + 4{z^2}} \right)} = \sqrt {1 - 4{z^2} + 4{z^3} - {z^4}}$

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.
f $$h\left( {z + k} \right)$$ Show Solution
$h\left( {z + k} \right) = \sqrt {1 - {{\left( {z + k} \right)}^2}} = \sqrt {1 - \left( {{z^2} + 2zk + {k^2}} \right)} = \sqrt {1 - {z^2} - 2zk - {k^2}}$