Domain & Range: Finding Functions And Cell Growth

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Hey guys! Let's break down how to find the domain and range of different functions, and then tackle a cell growth problem. We'll go through each function step-by-step, making sure everything is crystal clear.

Finding Domain and Range

a. f(x)=1βˆ’βˆ£xβˆ’3∣f(x) = 1 - |x - 3|

Okay, let's dive into our first function: f(x) = 1 - |x - 3|. The main goal here is to figure out what values we can plug in for x (that's the domain) and what values we'll get out for f(x) (that's the range).

Domain: For this function, we need to think about whether there are any restrictions on what x can be. Since it involves an absolute value, there are no restrictions! We can plug in any real number for x, and the function will give us a valid output. So, the domain is all real numbers.

In interval notation, the domain is: (βˆ’βˆž,∞)(-\infty, \infty).

Range: Now, let's figure out the range. The absolute value part, |x - 3|, will always be non-negative (zero or positive). So, the smallest it can be is 0. Since we are subtracting it from 1, the largest the function can be is when |x - 3| is 0. This happens when x = 3, and f(3) = 1 - |3 - 3| = 1 - 0 = 1. So, 1 is the maximum value.

As x moves away from 3 (either larger or smaller), |x - 3| gets bigger, and 1 - |x - 3| gets smaller. There's no lower limit, so the function can go all the way down to negative infinity.

In interval notation, the range is: (βˆ’βˆž,1](-\infty, 1].

To summarize, for f(x) = 1 - |x - 3|:

  • Domain: (βˆ’βˆž,∞)(-\infty, \infty)
  • Range: (βˆ’βˆž,1](-\infty, 1]

b. y=∣xβˆ’4∣xβˆ’4y = \frac{|x - 4|}{x - 4}

Next up, we have the function y = \frac{|x - 4|}{x - 4}. This one is a bit trickier because we have a fraction, and we need to avoid dividing by zero.

Domain: The denominator is x - 4, so we need to make sure that x - 4 β‰  0. This means x β‰  4. So, we can plug in any real number for x except for 4.

In interval notation, the domain is: (βˆ’βˆž,4)βˆͺ(4,∞)(-\infty, 4) \cup (4, \infty).

Range: Now for the range. Let's think about what happens when x > 4 and when x < 4.

  • If x > 4, then x - 4 is positive, so |x - 4| = x - 4. Therefore, y = \frac{x - 4}{x - 4} = 1.
  • If x < 4, then x - 4 is negative, so |x - 4| = -(x - 4). Therefore, y = \frac{-(x - 4)}{x - 4} = -1.

So, the function can only have two possible values: 1 or -1. Therefore, the range is just those two values.

In set notation, the range is: {-1, 1}.

To summarize, for y = \frac{|x - 4|}{x - 4}:

  • Domain: (βˆ’βˆž,4)βˆͺ(4,∞)(-\infty, 4) \cup (4, \infty)
  • Range: {-1, 1}

c. y=6xβˆ’3y = \frac{6}{x - 3}

Let's move on to the function y = \frac{6}{x - 3}. Again, we need to be careful about dividing by zero.

Domain: The denominator is x - 3, so we must ensure x - 3 β‰  0. This means x β‰  3. So, we can plug in any real number for x except for 3.

In interval notation, the domain is: (βˆ’βˆž,3)βˆͺ(3,∞)(-\infty, 3) \cup (3, \infty).

Range: Now, let's think about the range. As x gets very large (positive or negative), the fraction \frac{6}{x - 3} gets closer and closer to 0. However, it never actually equals 0 because the numerator is always 6.

Also, since we can make x as close to 3 as we want (but not equal to 3), we can make the fraction \frac{6}{x - 3} as large (positive or negative) as we want. So, the range is all real numbers except for 0.

In interval notation, the range is: (βˆ’βˆž,0)βˆͺ(0,∞)(-\infty, 0) \cup (0, \infty).

To summarize, for y = \frac{6}{x - 3}:

  • Domain: (βˆ’βˆž,3)βˆͺ(3,∞)(-\infty, 3) \cup (3, \infty)
  • Range: (βˆ’βˆž,0)βˆͺ(0,∞)(-\infty, 0) \cup (0, \infty)

d. y=x2βˆ’16y = \sqrt{x^2 - 16}

Lastly, we have the function y = \sqrt{x^2 - 16}. Here, we need to make sure that we're not taking the square root of a negative number, since we're dealing with real numbers.

Domain: We need xΒ² - 16 β‰₯ 0. This inequality can be factored as (x - 4)(x + 4) β‰₯ 0. To solve this, we can consider the intervals determined by the roots x = -4 and x = 4.

  • If x < -4, then both (x - 4) and (x + 4) are negative, so their product is positive.
  • If -4 < x < 4, then (x - 4) is negative and (x + 4) is positive, so their product is negative.
  • If x > 4, then both (x - 4) and (x + 4) are positive, so their product is positive.

So, the inequality is satisfied when x ≀ -4 or x β‰₯ 4.

In interval notation, the domain is: (βˆ’βˆž,βˆ’4]βˆͺ[4,∞)(-\infty, -4] \cup [4, \infty).

Range: Now for the range. Since we're taking the square root, the output will always be non-negative. The smallest value of xΒ² - 16 is 0 (when x = -4 or x = 4), so the smallest value of y is \sqrt{0} = 0. As x moves away from -4 and 4, xΒ² - 16 gets larger, so y also gets larger without bound.

In interval notation, the range is: [0,∞)[0, \infty).

To summarize, for y = \sqrt{x^2 - 16}:

  • Domain: (βˆ’βˆž,βˆ’4]βˆͺ[4,∞)(-\infty, -4] \cup [4, \infty)
  • Range: [0,∞)[0, \infty)

Cell Growth Problem

Now, let's shift gears and tackle the cell growth problem. We're given the function N(t) = 50000(1.6^{0.5t}), where N(t) is the number of cells at time t (in hours).

a. How many cells?

I assume you want to know how many cells are there at time t=0. Let's find the number of cells at t = 0. We plug in t = 0 into the function:

N(0) = 50000(1.6^{0.5 * 0}) = 50000(1.6^0) = 50000 * 1 = 50000

So, at t = 0, there are 50,000 cells.

Conclusion

Alright, guys! We've covered how to find the domain and range of different types of functions, including those with absolute values, fractions, and square roots. We've also solved a cell growth problem using an exponential function. I hope this was helpful and clear. Keep practicing, and you'll get the hang of it in no time!