# Defined amount and value amount of a function

What is the defined amount and value amount for this function:

$$f(x)=\sqrt{(x+7)(1-x)}?$$

The defined amount is all the x-values the function can be and the value amount is all the y-values the function can be but I don't know how to write it correctly.

The defined amount has to be $(x+7)(1-x) \ge 0$.

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What you are looking for is called the domain and the range of the function –  Hagen von Eitzen Apr 1 '13 at 10:29

See where function is not defined , those values of x are not in domain. So, domain is set of values of $x$ for which $f(x)$ is defined.

so,$$(x+7)(1-x)\ge0$$ which can we solved by wavy curve to get domain.

And range is the set of values $f(x)$ takes for all values of $x$ in domain.

$$f(x)=\sqrt{(x+7)(1-x)}$$

See here :http://goo.gl/JwRZF

See the real part of the plot. ie. in Domain=$[-7,1]$ the function acquires values from $0$ to $4$ to range=$[0,4]$

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The "defined amount" you seek, more commonly known as the domain of the function, is where the function is defined sensibly. Here we can assume we are looking for a subset of real numbers where the function is defined. The function is defined when the square root is defined, viz. when $(x+7)(1-x) \ge 0$.

The product of two real numbers is positive only when they both have same sign. So we can say the domain is either when $x+7 \ge 0 \text{ and } 1 - x \ge 0,$ or when $x+7 \le 0 \text{ and } 1 - x \le 0$.
$\quad x+7 \ge 0 \text{ and } 1 - x \ge 0 \implies -7 \le x \le 1$
$\quad x+7 \le 0 \text{ and } 1 - x \le 0 \implies 1 \le x \le -7$ which is not possible.
So the domain is D $= [-7, 1]$.

To find the "value amount", more commonly known as the range of the function, we look at possible values of $f(x)$, when $x \in$ D.

Now $f(x) = \sqrt{(x+7)(1-x)} = \sqrt{ 7 - 6x - x^2} = \sqrt{16 - (x+3)^2}$

The portion under the square root clearly has a maximum of $16$, so $f(x)$ has a maximum value of $4$, achieved when $x = -3$. As $x$ increases from this point, you could note that the function decreases, till it reaches $0$ when $x = 1$. Similarly as $x$ decreases from $x = -3$, the function decreases again, till it reaches $0$ when $x = -7$.

Thus the range of $f(x)$ is $[0, 4]$.

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