In Exercises 41 and 42, (a) write formulas for ƒ ○ g and g...

Question

In Exercises 41 and 42, (a) write formulas for ƒ ○ g and g ○ ƒ and find the (b) domain and (c) range of each.

ƒ(x) = 2 - x², g(x) = √ x + 2

Accepted Answer

Welcome back, everyone. Given the functions H X equals 3 x + 1 and K X equals X2 minus 4, write the formula for H K X and find its domain and range. So first of all, let's identify the composition. We're given our HX function. It is defined as a linear function 3 x + 1, and we also know that K of X is a parabolic function which is X2 minus 4. Let's understand that the compositionh of K X can be rewritten as H at a point k of x, even though it's not a point, it's a function. This allows us to visualize what we're trying to achieve. So basically it says that for every X that we have in H, we're going to replace it with K. For simplicity, we can say that H of K is going to be equal to 3, multiplied by K. We're replacing that X with K and adding 1. And now if we write H of K X. We're going to get 3 multiplied by k of X. Our k x is X2 minus 4, so we have performed the substitution, and then we are simply adding 1. So we have got our composition. What we want to do is simply expand it. If we distribute the terms, this gives us 3 X2 minus 12 + 1. So essentially minus 11, right? We have got the formula or H of K X. Let's write it down as our final answer. So that's 3 x2 minus 11. And now let's identify its domain and range. So for the domain, we have to understand that this is a polynomial function, right? A polynomial function is defined for all real numbers, so X belongs to the interval from negative infinity up to positive infinity. And now we have to consider its range. Well, it is a parabola that has a general form of AX2 + BX + C. And we have to recall that a parabola that has a leading coefficient that is positive. In this case, our leading coefficient is free. That parabola opens up, right? It extends towards infinity, so we can say that for our range we're going to have infinity. At the upper bound, and we want to identify the lower bound which corresponds to the minimum value at the vertex. Let's recall that we can identify the vertex using the formula XV equals negative B divided by 2A. Do we have any B coefficient? Well, we don't, right, because we don't have any BX term. We only have X squared, but there is no number multiplied by X. So our B is simply 1. Let's write that down. B B is 0, sorry. So our B is 0 and our A is the leading coefficient. It is equal to 3. So what we want to do is use this formula. We're going to get negative, 0 divided by 2 multiplied by 3, which gives us the X coordinate of 0. Now, from here, we want to identify the Y value, right? Because if we're looking at the range, well, our range corresponds to the range of values from the minimum Y to the maximum Y. So to identify the minimum Y value, we want to identify the Y coordinate at the vertex. So the Y value of the vertex is equal to. The value of the function, which is H of K X. Vertex, right, which is equal to h of K and our X of vertex is 0. So what we want to do is simply take our function 3x2 minus 11. And evaluate it at 0.0, so we're going to get 3 multiplied by 0 squared minus 11, which is -11. And this gives us the lower bound of the range. So the minimum value is -11, and we have to recall that it is inclusive, right, because this is the value for the vertex and our vertex is part of a parabola. So our range is Y belongs to the interval from -11 inclusive up to infinity, not inclusive. Well done, so let's label our answers. We first of all have our composition, which is 3 X2 minus 11, and then we have the domain. Which is all real numbers and the range which is Y belongs to the interval from -11 inclusive up to infinity, not inclusive. Thank you for watching.

In Exercises 41 and 42, (a) write formulas for ƒ ○ g and g ○ ƒ and find the (b) domain and (c) range of each.ƒ(x) = 2 - x², g(x) = √ x + 2

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In Exercises 41 and 42, (a) write formulas for ƒ ○ g and g ○ ƒ and find the (b) domain and (c) range of each.

ƒ(x) = 2 - x², g(x) = √ x + 2