f(x)= the absolute value of x^2-6x-7
a) Find a piecewise defined function that has no absolute value involved
b) Let g(x)=f(x) for x is greater than or equal to 7 find an expression of
g^-1(x)
c. give the domain and range of g^-1(x)
Since |x| = x when x >= 0
|x|= -x when x < 0
we need to find where f(x) >= 0 and f(x) < 0
f(x) = (x-7)(x+1)
f >= 0 where x <= -1 or x >= 7
This should get you going.
a) To find a piecewise defined function that has no absolute value involved, we need to consider the two cases:
Case 1: x^2 - 6x - 7 ≥ 0
In this case, the expression inside the absolute value, x^2 - 6x - 7, is greater than or equal to zero. We solve this inequality to find the interval where this holds true.
x^2 - 6x - 7 ≥ 0
(x - 7)(x + 1) ≥ 0
The critical points are x = -1 and x = 7. These divide the number line into three intervals: (-∞, -1], [-1, 7], [7, ∞].
Checking each interval:
1. For (-∞, -1]: In this interval, both factors are negative or zero, so the expression is positive. Therefore, x^2 - 6x - 7 ≥ 0 in this interval.
2. For [-1, 7]: In this interval, one factor is negative and one is positive, so the expression is negative. Therefore, x^2 - 6x - 7 < 0 in this interval.
3. For [7, ∞): In this interval, both factors are positive or zero, so the expression is positive. Therefore, x^2 - 6x - 7 ≥ 0 in this interval.
We can rewrite the function f(x) in terms of piecewise notation:
f(x) =
- (x^2 - 6x - 7), if x ≤ -1
(x^2 - 6x - 7), if -1 < x < 7
(x^2 - 6x - 7), if x ≥ 7
Note that we can simplify it to:
f(x) = (x^2 - 6x - 7), if x ≤ -1 or x ≥ 7
b) Let g(x) = f(x) for x ≥ 7. To find the expression for g^(-1)(x), we need to find the inverse of the function g(x).
Step 1: Replace g(x) with y: y = (x^2 - 6x - 7)
Step 2: Swap x and y: x = (y^2 - 6y - 7)
Step 3: Solve for y:
x = (y^2 - 6y - 7)
x = y^2 - 6y - 7
y^2 - 6y - (x + 7) = 0
Using the quadratic formula:
y = (6 ± √(6^2 - 4(1)(x + 7))) / (2(1))
y = (6 ± √(36 - 4x - 28)) / 2
y = (6 ± √(8 - 4x)) / 2
y = (6 ± 2√(2 - x)) / 2
y = 3 ± √(2 - x)
Therefore, the expression for g^(-1)(x) is g^(-1)(x) = 3 ± √(2 - x).
c. The domain and range of g^(-1)(x) can be determined by examining the domain and range of f(x). Since g^(-1)(x) is the inverse of g(x) = f(x), the domain and range of g^(-1)(x) will be the range and domain of f(x), respectively.
Domain of g^(-1)(x) = Range of f(x)
Since f(x) involves x^2, the range of f(x) is [0, ∞).
Range of g^(-1)(x) = Domain of f(x)
The domain of f(x) is all real numbers.
Therefore, the domain of g^(-1)(x) is all real numbers and the range of g^(-1)(x) is [0, ∞).
a) To find a piecewise defined function that has no absolute value involved, we need to consider the different cases where the absolute value function changes its behavior.
The given function is f(x) = |x^2 - 6x - 7|.
To remove the absolute value, we consider the cases where x^2 - 6x - 7 is positive and negative.
When x^2 - 6x - 7 < 0 (negative), we have f(x) = -(x^2 - 6x - 7).
When x^2 - 6x - 7 > 0 (positive), we have f(x) = x^2 - 6x - 7.
So, the piecewise defined function without absolute value can be written as:
f(x) = -(x^2 - 6x - 7), if x^2 - 6x - 7 < 0,
f(x) = x^2 - 6x - 7, if x^2 - 6x - 7 > 0.
b) Let g(x) = f(x) for x ≥ 7. This means that we take the function f(x) and only consider values of x that are greater than or equal to 7.
For this new function g(x), we replace x with x + 7 in f(x):
g(x) = f(x + 7)
= |(x + 7)^2 - 6(x + 7) - 7|
= |x^2 + 14x + 49 - 6x - 42 - 7|
= |x^2 + 8x|.
c) To find the expression of g^(-1)(x), we need to swap the roles of x and g(x) in the function g(x) and solve for x.
Let y = g(x), then x = g^(-1)(y). Rearranging the equation:
g(x) = y
x^2 + 8x = y.
Now, we solve for x:
x^2 + 8x - y = 0.
Using the quadratic formula, we get:
x = (-8 ± √(64 + 4y))/2
x = (-8 ± √(4(y + 16)))/2
x = -4 ± √(y + 16).
So, the expression for g^(-1)(x) is:
g^(-1)(x) = -4 ± √(x + 16).
c) The domain of g^(-1)(x) is all real numbers.
For the range of g^(-1)(x), consider that the square root term (√(x + 16)) is non-negative. So, the range of g^(-1)(x) is all real numbers greater than or equal to -4.