which function has the same range as $f(x) = -2\\sqrt{x - 3} + 8$?\n$g(x) = \\sqrt{x - 3} - 8$\n$g(x) =…
which function has the same range as $f(x) = -2\\sqrt{x - 3} + 8$?\n$g(x) = \\sqrt{x - 3} - 8$\n$g(x) = \\sqrt{x - 3} + 8$\n$g(x) = -\\sqrt{x - 3} - 8$\n$g(x) = -\\sqrt{x + 3} + 8$
Answer
Explanation:
Step1: Analyze the range of ( f(x) = -2\sqrt{x - 3} + 8 )
The square root function ( \sqrt{x - 3} ) has a range of ( [0, +\infty) ) because the square root of a non - negative number is non - negative. When we multiply ( \sqrt{x - 3} ) by ( - 2), the range of ( - 2\sqrt{x - 3} ) becomes ( (-\infty, 0] ) (since multiplying a non - negative number by a negative number reverses the inequality and scales it). Then, when we add 8 to ( - 2\sqrt{x - 3} ), the range of ( f(x)=-2\sqrt{x - 3}+8 ) is ( (-\infty, 8] ). The key property we are interested in is the sign of the coefficient of the square root term (negative) and the vertical shift. The horizontal shift (the ( x-3 ) inside the square root) does not affect the range, only the domain.
Step2: Analyze the range of each ( g(x) )
- For ( g(x)=\sqrt{x - 3}-8 ): The coefficient of the square root is positive. The range of ( \sqrt{x - 3} ) is ( [0,+\infty) ), so the range of ( \sqrt{x - 3}-8 ) is ( [-8,+\infty) ), which is different from the range of ( f(x) ).
- For ( g(x)=\sqrt{x - 3}+8 ): The coefficient of the square root is positive. The range of ( \sqrt{x - 3} ) is ( [0,+\infty) ), so the range of ( \sqrt{x - 3}+8 ) is ( [8,+\infty) ), which is different from the range of ( f(x) ).
- For ( g(x)=-\sqrt{x - 3}-8 ): The range of ( -\sqrt{x - 3} ) is ( (-\infty,0] ), and the range of ( -\sqrt{x - 3}-8 ) is ( (-\infty, - 8] ), which is different from the range of ( f(x) ).
- For ( g(x)=-\sqrt{x + 3}+8 ): The square root function ( \sqrt{x + 3} ) has a range of ( [0,+\infty) ). When we multiply by ( - 1 ), the range of ( -\sqrt{x + 3} ) is ( (-\infty,0] ). Then, when we add 8, the range of ( g(x)=-\sqrt{x + 3}+8 ) is ( (-\infty,8] ), which is the same as the range of ( f(x) ). The horizontal shift ( ( x + 3 ) instead of ( x - 3 )) affects the domain but not the range, and the negative coefficient of the square root and the vertical shift of 8 are the same in terms of determining the range as in ( f(x) ).