Suppose h(x) = x^2 + 8 x + 5, with the domain of h being the set of positive numbers. Evaluate h^{-1}(8).
for h(x)=x^2+8x+5
First reduce the quadratic expression to its standard form,
h(x)=(x+4)²-11
Find inverse of h(x):
1. interchange x and y
x=(y+4)²-11
2. solve for y
y=√(x+11)-4
Therefore
h-1(x) = √(x+11)-4
I'll leave it to you to find h-1(8) as an exercise.
As a check, plot both h(x) and h-1 on a graph. They should be symmetrical about the line y=x. Also note that the domain of h(x) is ℝ+, so the range of h-1 is also ℝ+.
To evaluate h^(-1)(8), we need to find the value of x that makes h(x) equal to 8.
First, let's write the equation h(x) = 8:
x^2 + 8x + 5 = 8
Rearranging the equation, we get:
x^2 + 8x - 3 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 8, and c = -3. Substituting the values into the quadratic formula:
x = (-8 ± sqrt(8^2 - 4*1*(-3))) / (2*1)
Simplifying further:
x = (-8 ± sqrt(64 + 12)) / 2
x = (-8 ± sqrt(76)) / 2
x = (-8 ± sqrt(4 * 19)) / 2
x = (-8 ± 2sqrt(19)) / 2
Now we have two possible values for x:
x₁ = (-8 + 2sqrt(19)) / 2
x₁ = -4 + sqrt(19)
x₂ = (-8 - 2sqrt(19)) / 2
x₂ = -4 - sqrt(19)
However, we need to consider only the positive numbers as the domain of h. Therefore, the only valid solution is:
x = -4 + sqrt(19)
So, h^(-1)(8) = -4 + sqrt(19).