![Generate an image that illustrates the abstract concept of two mathematical functions and their combined outcome. Display the following elements without text : Two distinct curves to represent function f and g individually. Function f, represented as a positively sloped straight line starting at [0,3] to denote the equation 2x + 3 for x ≥ 0. Function g, should be a parabola starting from an unknown point q and extending to negative infinity. Finally, incorporate a third curve to represent the function fg which should be a downward opening parabola with the equation 6x^2 - 21 for x ≤ q. Note: avoid adding any labels, numbers or text to the image.](/images/questions/1504144/1024x1024.png)
The function f is such that f(x)�
= 2x + 3 for x ≥ 0.
The function g is such that g(x)�= ax^2 + b for x ≤ q, where a, b and q are constants. The function fg is such that fg(x)�= 6x^2 − 21 for x ≤ q.
i)Find the values of a and b
ii) Find the greatest possible value of q.
It is now given that q = -3
iii) Find the range of fg
iv)Find the expression for inverse of (fg)(x) ad state the domain of the inverse of (fg)
i) To find the values of a and b, we can set up equations using the given information:
Since g(x) = ax^2 + b for x ≤ q, we can substitute x = q into g(x) to obtain g(q) = aq^2 + b.
Since fg(x) = 6x^2 - 21 for x ≤ q, we can substitute x = q into fg(x) to obtain fg(q) = 6q^2 - 21.
Since fg(q) = g(q), we can equate the two expressions and solve for a and b:
6q^2 - 21 = aq^2 + b
Comparing the coefficients of q^2, we get:
6 = a
Comparing the constant terms, we get:
-21 = b
Therefore, the values of a and b are 6 and -21, respectively.
ii) The greatest possible value of q is given as q = -3.
iii) To find the range of fg, we need to consider the values of fg(x) for x ≤ q.
Since fg(x) = 6x^2 - 21, we can substitute x = q into fg(x) to obtain fg(q) = 6q^2 - 21.
Since we are looking at values of x ≤ q, we substitute x = q - δ (a small positive number) into fg(x). This will give us a decreasing function as x approaches q.
Therefore, the range of fg is (-∞, 6q^2 - 21] or (-∞, -63].
iv) To find the expression for the inverse of fg(x), we can switch the roles of x and y and solve for y.
Let y = fg(x) = 6x^2 - 21.
Swap x and y: x = 6y^2 - 21.
Rearrange the equation: 6y^2 = x + 21.
Divide both sides by 6: y^2 = (x + 21) / 6.
Take the square root of both sides (considering the positive square root only): y = √[(x + 21) / 6].
Therefore, the expression for the inverse of fg(x) is given by fg^(-1)(x) = √[(x + 21) / 6], and the domain of the inverse of fg is (-∞, -21].
i) To find the values of a and b, we need to make use of the given information about the function fg(x).
Given that fg(x) = 6x^2 - 21 for x ≤ q, we can substitute x = q into fg(x) and equate it to g(x) to form an equation:
6q^2 - 21 = aq^2 + bq
From here, we have one equation with two unknowns (a and b). To solve for a and b, we need another equation involving a and b. We can make use of the information about the function f(x).
Given that f(x) = 2x + 3 for x ≥ 0, we can substitute x = 0 into f(x) and equate it to g(x) for x = q:
2(0) + 3 = a(0)^2 + bq
3 = bq
Now we have two equations:
1) 6q^2 - 21 = aq^2 + bq
2) 3 = bq
Solving these two equations simultaneously will give us the values of a and b.
ii) To find the greatest possible value of q, given that q = -3, we need to consider the domain of the function fg(x). The function fg(x) is defined as fg(x) = 6x^2 - 21 for x ≤ q. In this case, x ≤ q means that x cannot be greater than q. Therefore, the greatest possible value of q is the largest value that x can take while still satisfying the condition x ≤ q. In this case, q = -3, so the greatest possible value of q is -3.
iii) To find the range of fg, we need to determine the set of all possible y-values that fg(x) can take. From the equation fg(x) = 6x^2 - 21, we can observe that the coefficient of the x^2 term is positive (6), which means that the parabola opens upward. This implies that fg(x) can take on any value greater than or equal to the vertex of the parabola. The vertex of the parabola can be determined by finding the x-coordinate of the vertex using the formula x = -b/2a.
For fg(x) = 6x^2 - 21, a = 6 and b = 0. Substituting these values into the formula x = -b/2a, we get:
x = -(0)/(2(6)) = 0
Therefore, the vertex of the parabola is (0, -21).
Since the parabola opens upward and the vertex is the lowest point on the graph, the range of fg is (-21, ∞).
iv) To find the expression for the inverse of (fg)(x) and state its domain, we need to switch the roles of x and y in the equation fg(x) = 6x^2 - 21 and solve for x.
Start with fg(x) = 6x^2 - 21:
y = 6x^2 - 21
Switching x and y:
x = 6y^2 - 21
Now solve for y:
6y^2 = x + 21
y^2 = (x + 21)/6
y = ±sqrt((x + 21)/6)
The expression for the inverse of (fg)(x) is y = ±sqrt((x + 21)/6).
For the domain of the inverse of (fg), we need to consider the values of x that make the square root expression defined. Since the square root function is only defined for non-negative values, the expression inside the square root, (x + 21)/6, must be greater than or equal to 0:
(x + 21)/6 ≥ 0
Solving this inequality:
x + 21 ≥ 0
x ≥ -21
Therefore, the domain of the inverse of (fg) is x ≥ -21.
Thanks a lot Steve!!!
i) To find the values of a and b, we need to equate fg(x) and g(x) and solve for a and b.
Since fg(x) = 6x^2 - 21 and g(x) = ax^2 + b, we have:
6x^2 - 21 = ax^2 + b
Comparing the coefficients of x^2, we get:
6 = a
Comparing the constant terms, we get:
-21 = b
Therefore, a = 6 and b = -21.
ii) The greatest possible value of q can be found by finding the value of x such that fg(x) is defined. Since q is given as -3, we need to ensure that the domain of fg is valid for x ≤ q.
fg(x) = 6x^2 - 21, where x ≤ q
Since q = -3, we have:
6x^2 - 21 is defined for x ≤ -3
Therefore, the greatest possible value of q is -3.
iii) The range of fg can be determined by analyzing the function fg(x) = 6x^2 - 21.
The range of fg is all possible values that fg(x) can take. In this case, since fg(x) is a quadratic equation of the form 6x^2 - 21, the range is all real numbers.
So, the range of fg is (-∞, +∞), which means fg(x) can take any real value.
iv) To find the expression for the inverse of fg, we need to switch the roles of x and y in the equation fg(x) = 6x^2 - 21 and solve for x.
Let y = 6x^2 - 21, then the equation becomes:
x = 6y^2 - 21
Rearranging this equation to solve for y, we have:
6y^2 = x + 21
Dividing both sides by 6, we get:
y^2 = (x + 21)/6
Taking the square root of both sides, we have:
y = ±√((x + 21)/6)
Therefore, the expression for the inverse of fg(x) is:
(fg)^-1(x) = ±√((x + 21)/6)
The domain of the inverse of fg is all real numbers, since the square root function is defined for all real numbers.
fg = (2x+3)(ax^2+b) = 6x^2-21
2ax^3+3ax^2+bx+3b = 6x^2-21
when x=q, we have
2aq^3+3aq^2+bq+3b=6q^2-21
2aq^3+(3a-6)q^2+bq+(3b-21) = 0
check out the solutions to that at this url. The only easy one is in fact when q = -3.
http://www.wolframalpha.com/input/?i=2aq%5E3%2B(3a-6)q%5E2%2Bbq%2B(3b-21)+%3D+0