I'm working on an optional test review packet from my Precalculus teacher and out of the whole packet, only these three questions stumped me:
1. Find the range of the function 2^(x^2 + 4).
2. Create a function that requires the given domain:
a) [5,0)
b) -4 ≤ x < -1 or 1 < x ≤ 4
Please help me understand how I would find the answer to the questions. Thanks!
I've just figured out question 1. Can someone please help me on question 2, parts a) and b), ASAP?
I'm confused as well. You can restrict any function's domain just by defining what it is, as in a piecewise function.
Ah ok, thanks!
To find the range of a function, you need to determine the set of all possible outputs (or y-values) that the function can take. For the given function, 2^(x^2 + 4), you can approach it by analyzing the properties of exponential functions and using some algebraic manipulation.
1. Find the range of the function 2^(x^2 + 4):
To determine the range, we have to consider what happens to the function as x varies over all possible values.
First, note that the base of the exponential function, 2, is positive and greater than 1. This means that the function is always increasing as x increases.
Next, consider the exponent term x^2 + 4. Since x^2 is always non-negative (or positive) for any real value of x, and adding 4 to it doesn't change that fact, the exponent is always non-negative. Therefore, the function is always positive or zero.
Combining these observations, we can conclude that the range of the function 2^(x^2 + 4) is the set of positive real numbers, including zero. In interval notation, this can be written as [0, +∞).
2. Create a function that requires the given domain:
a) [5,0):
To create a function that satisfies the given domain [5,0), you need to consider the properties of linear functions. A linear function has the form y = mx + b, where m is the slope and b is the y-intercept.
To satisfy the given domain, the function should start from x = 5 and end at x = 0. So, we can set the y-intercept as the starting point (5, y) and find the slope by considering the change in y over the change in x.
Let's assume the y-intercept is (0, c), where c is any real number. Then the slope of the linear function would be (5 - 0) / (0 - 5) = -1.
Therefore, a possible linear function that satisfies the given domain is y = -x + c, where c is any real number.
b) -4 ≤ x < -1 or 1 < x ≤ 4:
To create a function that satisfies this domain, you need to consider the properties of piecewise-defined functions. A piecewise function has different definitions for different intervals of the domain.
For the given domain, you can divide it into two intervals: -4 ≤ x < -1 and 1 < x ≤ 4.
For the first interval, -4 ≤ x < -1, you can use a linear function with a positive slope.
Let's assume the function for this interval is y = mx + b, where m is the slope and b is the y-intercept. Since we need a positive slope, we can choose m = 2, for example. To determine the y-intercept b, we can use the point (-1, y), where y can be any real number.
For the second interval, 1 < x ≤ 4, you can use another linear function, but with a different slope and y-intercept. You can choose a negative slope, for example, m = -2. Again, determine the y-intercept using a point (1, y), where y can be any real number.
Putting it all together, a possible function that satisfies the given domain is:
-4 ≤ x < -1: y = 2x + b, where b is any real number.
1 < x ≤ 4: y = -2x + c, where c is any real number.
Note that there are multiple valid functions that can satisfy a given domain, so these are just examples.
I hope this helps you understand how to find the answers to these questions! Let me know if you have any further doubts or need additional clarification.