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Find f²⁰²³ (x) if f is

i) f(x)=x² −x+3
ii) f(x)=x²⁰²³ −3x²⁰²² +5x² +10
iii) f(x)= 1/x


User Saran
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Final Answer:

i) f²⁰²³(x) for f(x) = x² - x + 3 is a complex expression involving high powers of x, which when raised to the power of 2023 becomes intricate to calculate explicitly without further context or patterns.

ii) f²⁰²³(x) for f(x) = x²⁰²³ - 3x²⁰²² + 5x² + 10 is similarly complex, involving exceedingly high powers of x, making direct computation infeasible within the given context.

iii) f²⁰²³(x) for f(x) = 1/x yields f²⁰²³(x) = x⁻²⁰²³, signifying that the function raised to the power of 2023 results in x to the power of negative 2023.

Step-by-step explanation:

In the cases of polynomials like f(x) = x² - x + 3 and f(x) = x²⁰²³ - 3x²⁰²² + 5x² + 10, finding f²⁰²³(x) involves an exceedingly high power of x, making direct computation impractical within a given context without specific patterns or additional information. However, for f(x) = x² - x + 3, the function is a quadratic polynomial, and raising it to the power of 2023 would result in a high-degree polynomial, which requires extensive computation and analysis, lacking a straightforward calculation method.

For the function f(x) = 1/x, raising it to the power of 2023 results in f²⁰²³(x) = x⁻²⁰²³. In this case, the negative exponent signifies that the function raised to a high power leads to x raised to the power of negative 2023. This transformation highlights the inverse relationship and indicates that the function raised to a high power yields the reciprocal of x raised to a significant negative power, showcasing the impact of exponentiation on functions involving reciprocals. The process of raising functions to high powers, especially when dealing with inverses like 1/x, demonstrates the notable effects of exponentiation on various functions.

User Rampatowl
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