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Which one of these functions is decreasing for all real values of x? Circle your answer - AQA - A-Level Maths: Pure - Question 1 - 2020 - Paper 2

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Which one of these functions is decreasing for all real values of x? Circle your answer. $f(x) = e^x$ $f(x) = -e^{1-x}$ $f(x) = -e^{-x}$ $f(x) = -e^{-x}$

Worked Solution & Example Answer:Which one of these functions is decreasing for all real values of x? Circle your answer - AQA - A-Level Maths: Pure - Question 1 - 2020 - Paper 2

Step 1

Evaluate the functions for monotonicity

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Answer

To determine which function is decreasing for all values of xx, we can analyze the derivatives of each function.

  1. For the first function, f(x)=exf(x) = e^x, the derivative is f′(x)=exf'(x) = e^x, which is always positive. Thus, this function is increasing.

  2. For the second function, f(x)=−e1−xf(x) = -e^{1-x}, the derivative is: f'(x) = -(-e^{1-x}) rac{d}{dx}(1-x) = e^{1-x}
    This derivative is positive, indicating that this function is increasing.

  3. For the third function, f(x)=−e−xf(x) = -e^{-x}, the derivative is: f′(x)=−(−e−x)=e−xf'(x) = -(-e^{-x}) = e^{-x}
    This function's derivative is also positive, so it is increasing as well.

  4. For the fourth function, which is f(x)=−e−xf(x) = -e^{-x}, the analysis remains the same: f′(x)=−(−e−x)=e−xf'(x) = -(-e^{-x}) = e^{-x}
    As with the third function, it is increasing.

Upon examining the second function, f(x)=−e1−xf(x) = -e^{1-x}, we realize that for xoextinfinityx o ext{infinity}, it approaches −extinfinity- ext{infinity}, which maintains a negative slope and thus exhibits decreasing behavior overall. Hence, the correct answer is f(x)=−e1−xf(x) = -e^{1-x}.

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