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Chapter 10: Problem 857

In which interval \(\mathrm{f}(\mathrm{x})=\sin ^{4} \mathrm{x}+\cos ^{4} \mathrm{x}, \mathrm{x} \in[0,(\pi / 2)]\) is increasing function (a) \(|(\pi / 4),(\pi / 2)|\) (b) \(|0,(\pi / 4)|\) (c) \(|0,(\pi / 2)|\) (d) \(|(\pi / 4),\\{(-\pi) / 2\\}|\)

Short Answer

Expert verified
The function \(f(x) = \sin^4(x) + \cos^4(x)\) is increasing on the interval \(x \in [0, (\pi/4)]\), which corresponds to answer choice (b).

Step by step solution

01

Find the derivative of the function

To find the derivative of the function f(x) = sin^4(x) + cos^4(x), we can use the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. First, let's find the derivatives of sin^4(x) and cos^4(x): Using the chain rule, we get: \(\frac{d}{dx}(\sin^4(x)) = 4\sin^3(x)\cos(x)\) (where we used the fact that \(\frac{d}{dx}(\sin(x)) = \cos(x)\)) \(\frac{d}{dx}(\cos^4(x)) = -4\cos^3(x)\sin(x)\) (where we used the fact that \(\frac{d}{dx}(\cos(x)) = -\sin(x)\)) Then, we add the two derivatives to obtain the derivative of f(x): \(\frac{df}{dx} = 4\sin^3(x)\cos(x) - 4\cos^3(x)\sin(x)\)
02

Simplify the derivative

To simplify the derivative, factor out 4sin(x)cos(x): \(\frac{df}{dx} = 4\sin(x)\cos(x)(\sin^2(x) - \cos^2(x))\)
03

Find the intervals where the derivative is positive

For a function to be increasing, the derivative must be greater than 0. We want to find the interval on which \(\frac{df}{dx} > 0\). Notice that: 1. \(\sin^2(x) - \cos^2(x)\) will be positive when sin(x) > cos(x), which is true between [0, π/4] 2. \(\sin^2(x) - \cos^2(x)\) will be negative when sin(x) < cos(x), which is true between [π/4, π/2] Thus, when x is in [0, π/4], \(\frac{df}{dx} > 0\), and when x is in [π/4, π/2], \(\frac{df}{dx} < 0\).
04

Choose the correct answer

So, f(x) = sin^4(x) + cos^4(x) is an increasing function on the interval x ∈ [0, π/4], which corresponds to answer choice (b).

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