ƒ¹(x) [g(x) ± h(x)]
To answer this question, we need to understand the notation being used
To answer this question, we need to understand the notation being used. The expression “ƒ¹(x)” refers to the inverse function of ƒ(x). The inverse function is obtained by swapping the roles of x and ƒ(x), meaning that if ƒ(a) = b, then ƒ¹(b) = a.
Now, let’s consider the expression “ƒ¹(x) [g(x) ± h(x)]”. In this expression, we have three functions: ƒ¹(x), g(x), and h(x). The square brackets represent the application of the function ƒ¹(x) on the expression inside. The ± symbol is used to indicate that we can either add or subtract the functions g(x) and h(x).
To evaluate this expression or simplify it further, we would need more information about the specific functions ƒ(x), g(x), and h(x). Without knowing the actual functions, it is challenging to provide a specific answer. However, I can explain the general idea of how to approach similar expressions.
1. Evaluate g(x) ± h(x):
First, calculate g(x) and h(x) independently. Let’s say g(x) = 2x and h(x) = 3x^2. The expression then becomes:
ƒ¹(x) [2x ± 3x^2]
2. Apply the inverse function ƒ¹(x):
Plug in the value of x into the inverse function ƒ¹(x). Let’s assume ƒ¹(x) = x^2. The expression becomes:
(x^2) [2x ± 3x^2]
3. Simplify further:
Distribute the terms inside the square brackets:
(2x^3 ± 3x^4)
Again, this simplification is based on assumed functions for illustration purposes. You would need to know the actual functions ƒ(x), g(x), and h(x) to provide a more accurate evaluation or simplification of the expression.
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