determine the open intervals on which the function is increasing, decreasing, or constant. (enter your…

determine the open intervals on which the function is increasing, decreasing, or constant. (enter your answers using interval notation. if an answer does not exist, enter dne.)\n\nf(x) = \\sqrt{x^{2}-25}\n\nincreasing\n\nconstant\ndecreasing

determine the open intervals on which the function is increasing, decreasing, or constant. (enter your answers using interval notation. if an answer does not exist, enter dne.)\n\nf(x) = \\sqrt{x^{2}-25}\n\nincreasing\n\nconstant\ndecreasing

Answer

Explanation:

Step1: Analyze the function's domain

The function ( f(x)=\sqrt{x^{2}-25} ) is defined when ( x^{2}-25\geq0 ), i.e., ( x\leq - 5 ) or ( x\geq5 ).

Step2: Use the derivative to find intervals

First, find the derivative using the chain - rule. Let ( u = x^{2}-25 ), then ( y=\sqrt{u}=u^{\frac{1}{2}} ). The derivative ( y^\prime=\frac{dy}{du}\cdot\frac{du}{dx} ). ( \frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{x^{2}-25}} ) and ( \frac{du}{dx} = 2x ). So ( f^\prime(x)=\frac{x}{\sqrt{x^{2}-25}} ).

  • For the increasing interval: Set ( f^\prime(x)>0 ). Since ( \sqrt{x^{2}-25}>0 ) for ( x\neq\pm5 ), then ( x>0 ). Combining with the domain ( x\geq5 ), the increasing interval is ( [5,\infty) ).
  • For the decreasing interval: Set ( f^\prime(x)<0 ). Since ( \sqrt{x^{2}-25}>0 ) for ( x\neq\pm5 ), then ( x < 0 ). Combining with the domain ( x\leq - 5 ), the decreasing interval is ( (-\infty,-5] ).

Answer:

Increasing: ( [5,\infty) ); Decreasing: ( (-\infty,-5] ); Constant: DNE