Question #ff602

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Question #ff602

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Answer 1 · 2017-11-29T08:15:46

Range in interval notation is: $(- \infty, 0] \cup (1, \infty)$ $(-oo, 0] uu (1, oo)$
or you can say: $y \leq 0 or y > 1$ $y <=0 " or " y > 1$

Explanation:

Given: $y = f (x) = \frac{x + 3}{4 - \sqrt{x^{2} - 9}}$ $y = f(x) = (x+3)/(4-sqrt(x^2-9))$

The range is the valid values of $y$ $y$ , which depend on the values of $x$ $x$ (the domain).

To find the domain, you must take into consideration both the. square root (values must be $\geq 0$ $>= 0$ ) and the fact that the denominator must not be $= 0$ $= 0$ .

Square root:
Difference of squares: $x^{2} - 9 \geq 0$ $x^2-9 >=0$

$x^{2} - 3^{2} \geq 0$ $x^2 - 3^2 >=0$

When $(x - 3) (x + 3) = 0, x = \pm 3$ $(x-3)(x+3) =0, x = +-3$

When $x \leq - 3 and x \geq 3$ $x<= -3 " and " x>= 3$ then $x^{2} - 9 \geq 0$ $x^2-9 >=0$

Denominator:
$4 - \sqrt{x^{2} - 9} \neq 0$ $4-sqrt(x^2-9) != 0$

$4 \neq \sqrt{x^{2} - 9}$ $4 != sqrt(x^2-9)$

Square both sides of the equation:
$16 \neq x^{2} - 9$ $16 != x^2-9$

$16 + 9 \neq x^{2}$ $16 + 9 != x^2$

$25 \neq x^{2}$ $25 != x^2$

$x \neq - 5 and x \neq 5$ $x != -5 " and " x!= 5$
This means there are vertical asymptotes at these values.

Domain in interval notation is
$(- \infty, - 5) \cup (- 5, - 3] \cup [3, 5) \cup (5, \infty)$ $(-oo, -5) uu (-5, -3] uu [3, 5) uu (5, oo)$

To find the range, you need to find out if there are any horizontal asymptotes. The simplest way is to input several large negative numbers and several large positive numbers into the function to see if the value of $y$ $y$ converges:

Let $x = - 50, y = 1.024$ $x = -50, y = 1.024$
Let $x = - 100, y = 1.01$ $x = -100, y = 1.01$
Let $x = 50, y = - 1.15$ $x = 50, y = -1.15$
Let $x = 100, y = - 1.07$ $x = 100, y = -1.07$
Let $x = 1000, y = - 1.007$ $x = 1000, y = -1.007$

We have a horizontal asymptote at $y = 1 and y = - 1$ $y = 1 " and " y = -1$

If we just look at this information, the Range is $(- \infty, - 1) \cup (1, \infty)$ $(-oo, -1)uu(1, oo)$

But we must be careful, because the horizontal asymptotes aren't required to exist along the whole function, only on the far left and far right. This means we must also check the values at each of the domains limiting values:

Let $x = - 5, y = undefined$ $x = -5, y = "undefined"$
Let $x = - 3, y = 0$ $x = -3, y = 0$
Let $x = 3, y = 1.5$ $x = 3, y = 1.5$
Let $x = 5, y undefined$ $x = 5, y "undefined"$

From this information we see that when $x = - 3, y = 0$ $x = -3, y = 0$ .

This means the range is actually: $(- \infty, 0] \cup (1, \infty)$ $(-oo, 0] uu (1, oo)$

or you can say the range is $y \leq 0 or y > 1$ $y <=0 " or " y > 1$

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Question #ff602

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