Composition of functions occurs when we have two functions normally written similar or exactly like f(x) & g(x) - you can have any coefficients to the (x), but the most commonly seen are f(x) and g(x). They are written as either f(g(x)) or (f o g)(x). Because our composition is written as $(f \circ g)(2)$ , we are replacing the x values in the g(x) function with 2 and simplifying the expression.

$g(2) = 3(2)$

$g(2) = 6$

Now, because we are composing the functions, this value we have solved for now replaces the x-values in the f(x) function. So, f(x) becomes f(6), and we use the same manner as above to simplify.

$f(6) = (6)^2-2$

$f(6) = 36-2$

$f(6) = 34$

Therefore, when we compose the functions, our final answer is $\bold{f(6) = 34}$ .

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3 years ago

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In the xy, y-plane, the parabola with equation y = ( x− 11)^2y=(x−11) 2 y, equals, (, x, −, 11, ), squaredintersects the line wi

IRINA_888 [86]

Answer:

10 units

Step-by-step explanation:

Allow me to revise your question for a better understanding.

<em>"In the xy-plane, the parabola with equation y = (x − 11) ² intersects the line with equation y = 25 at two points, A and B. What is the length of AB" </em>

Here is my answer

Because the parabola intersects the line with equation y = 25 Substituting y = 25 in the equation of the parabola y = (x - 11)², we get

25 = (x - 11)²

<=>x - 11 = ± 5

<=> $\left \{ {{x=6} \atop {x=16}} \right.$

Thus A(16, 25) and B(6, 25) are the points of intersection of the given parabola and the given line.

So the length of AB = √[(16 - 6)² + (25 - 25)²]

= √100 = 10 units

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3 years ago

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