Mathematics: Let y=f(x)=(x^2-8 sqrt(-x))^2. find the intervals on which f is increasing/decreasing

Let y=f(x)=(x^2-8 sqrt(-x))^2. find the intervals

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miguelc2145

20.07.2019

Answer on assumption that f is dealt in real numbersf is increasing on the open interval (-4, -∛4) and f is decreasing on the open intervals (-∞, -4) and (-∛4, 0).======assuming the domain of x cannot contain complex numbers, we establish first that the domain is x ≤ 0 due to the presence of √(-x).to find where f is increasing/decreasing, we must find where the derivative of f is positive or negative. in order to do that, we must differentiate: $\begin{aligned} f(x) & = \left(x^2-8\sqrt{-x} \right)^2 \\ f'(x) & = \left[ \left(x^2-8\sqrt{-x} \right)^2 \right]'\\ & = 2\left(x^2-8\sqrt{-x} \right) \cdot \left(x^2-8\sqrt{-x} \right)' & & \text{\footnotesize chain rule} \\ & = 2\left(x^2 - 8\sqrt{-x} \right) \left(2x - 8 \cdot \frac{1}{2\sqrt{-x}} \cdot -1 \right) \\ & = 2\left(x^2 - 8\sqrt{-x} \right) \left( 2x + \frac{4}{\sqrt{-x} } \right) \end{aligned}$ note that $\dfrac{d}{dx} \sqrt{-x} = \dfrac{d}{dx} (-x)^{1/2} = -\frac{1}{2} (-x)^{-1/2} \cdot (-1 )$ find the critical numbers. these are where f' is undefined or equal to zero.we can see immediately that f' is undefined at x = 0 since there would be a division by zero.x = 0 is a critical number.to find where f' is equal to zero, set f'(x) = 0 and solve. in this case, we can use the zero factor property. $\displaystyle 0 = 2\left(x^2 - 8\sqrt{-x} \right) \left( 2x + \frac{4}{\sqrt{-x} } \right)$ setting $x^2 - 8\sqrt{-x}$ to equal zero: $\begin{aligned} x^2 - 8\sqrt{-x} & = 0 \\ x^2 & = 8\sqrt{-x} \\ \left( x^2 \right)^2 & = \left( 8\sqrt{-x} \right)^2 \\ x^4 & = 64(-x) \\ x^3 & = -64 \\ x & = \sqrt[3]{-64} = - 4 \end{aligned}$ so x = -4 is a critical number.setting $2x + \frac{4}{\sqrt{-x}}$ to equal zero: $\begin{aligned} 2x + \frac{4}{\sqrt{-x}} & = 0 \\ 2x & = -\frac{4}{\sqrt{-x}} \\ x & = -\frac{2}{\sqrt{-x}} \\ \left( x \right)^2 & =\left( -\frac{2}{\sqrt{-x}} \right)^2 \\ x^2 & = \frac{4}{-x} \\ x^3 & = -4 \\ x & = \sqrt[3]{-4} = -\sqrt[3]{4} \end{aligned}$ so x = -∛(4) is a critical number.our critical numbers are -4, -∛(4), and 0. setting up an interval chart to test for the sign of f' in the intervals created by these critical numbers, plug in a number that is in each interval to find the sign. $\begin{array}{l | c | r} \text{interval} & f'(x) & f \text{ inc/dec} \\ (-\infty, -4) & - & \text{decreasing} \\ (-4, -\sqrt[3]{4} ) & + & \text{increasing} \\ (-\sqrt[3]{4}, 0) & - & \text{decreasing} \\ (0, \infty) & n/a & \text{not real}\end{array}$ therefore, f is increasing on the open interval (-4, -∛4) and f is decreasing on the open intervals (-∞, -4) and (-∛4, 0).attached is a graph of function f from desmos to confirm

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Let y=f(x)=(x^2-8 sqrt(-x))^2. find the intervals on which f is increasing/decreasing

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