I am to write this in interval notation. I don't remember learning this:
But here are two examples. On a graph, i know how to get the domain and range, but not from a function.
f(x) = -4 -the radical(9-(x^2))
g(x) = SIN the radical(x^2(-1)
I am to write this in interval notation. I don't remember learning this:
But here are two examples. On a graph, i know how to get the domain and range, but not from a function.
f(x) = -4 -the radical(9-(x^2))
g(x) = SIN the radical(x^2(-1)
Recall that the domain of a function is all possibilities of x at which f(x) is defined:
$\displaystyle f(x)=-4-\sqrt{9-x^2}$
where is this function defined? Well, you know that you can't take the sqrt of a negative number, s f(x) is defined every where that $\displaystyle 9-x^2\geq0\Rightarrow\mid{x}\mid\leq{3}$. (If x where any bigger than 3, 9-x^2 would be less than zero. A big no no unless we're considering complex numbers.)
In interval notation: $\displaystyle [-3,3]$
you can do the other now can't you?
Solve for x and then determine what y cannot be:
$\displaystyle y=-4-\sqrt{9-x^2}$
$\displaystyle y+4=-\sqrt{9-x^2}$
$\displaystyle (-y-4)^2=9-x^2$
$\displaystyle (-y-4)^2-9=-x^2$
$\displaystyle \pm\sqrt{-y^2-8y-7}=x$
$\displaystyle \pm\sqrt{-[(y+7)(y+1)]}=x$
This one requires alittle more thought..........
Ask yourself:
When is the expression under the radical sign negative?
For range I usually look at its graph, or plug values of x into my head.
$\displaystyle y=-4-\sqrt{9-x^2}$
The domain is [-3, 3]. If you look at $\displaystyle \sqrt{9-x^2}$ only, what is its minimum value? It's 0, when x is -3 or 3. The maximum value? It's 3, when x = 0.
Now consider the whole function. If you plug in x = -3 or 3, you actually get the maximum value of the function, because there is a negative in front of the square root:
$\displaystyle y=-4-\sqrt{9-9} = -4$
Plug in 0 for x to get the minimum value of the function:
$\displaystyle y=-4-\sqrt{9-0} = -7$
The range is [-7, -4].
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