SOLUTION: find range ? {{{sqrt(-x^2+4x-3)+sqrt(sin(pi/2*sin(pi/2)(x-1)))))}}}

Algebra ->  Functions -> SOLUTION: find range ? {{{sqrt(-x^2+4x-3)+sqrt(sin(pi/2*sin(pi/2)(x-1)))))}}}      Log On


   

Question 696709: find range ?
sqrt%28-x%5E2%2B4x-3%29%2Bsqrt%28sin%28pi%2F2%2Asin%28pi%2F2%29%28x-1%29%29%29%29%29
Answer by mouk(232) About Me  (Show Source):
You can put this solution on YOUR website!
Let
+y+=+sqrt%28-x%5E2%2B4x-3%29+%2B+sqrt%28sin%28pi%2F2%2Asin%28pi%2F2%29%28x-1%29+%29%29+

Then factorising the first term and tidying up the second gives:
+y+=+sqrt%28%28-x%2B1%29%28x-3%29%29+%2B+sqrt%28sin%28pi%2F2%2A%28x-1%29+%29%29+


As we are assuming a real-valued function then both terms under the square root signs must be simultaneously positive (or zero).


Firstly consider +%28-x%2B1%29%28x-3%29+%3E=+0+
+graph%28+200%2C+200%2C+-1%2C+4%2C+-10%2C+10%2C+-x%5E2%2B4x-3+%29+
The solution to this is: +1+%3C=+x+%3C=3+


Secondly, consider +sin%28pi%2F2%2A%28x-1%29%29+%3E=+0
+graph%28+200%2C+200%2C+-2pi%2C+2pi%2C+-3%2C+3%2C+sin%28pi%2F2%2A%28x-1%29%29+%29+
+sin%28pi%2F2%2A%28x-1%29%29+%3E=+0+
+0%3C=+pi%2F2%2A%28x-1%29+%3C=+pi+
+0%3C=+x-1+%3C=+2+
+1%3C=+x+%3C=+3+


So in both cases, the square root terms simultaneously exist when +1%3C=+x+%3C=+3+, and this is the range.