Solve the inequality sinx>square root x.
sin x > √x
we can say immediately that x≥0 , since we can't take the square root of a negative.
so I would graph y = sinx
and y = √x and look at the region where the sinx is above the √x
http://www.wolframalpha.com/input/?i=y+%3D+sin%28x%29+%2C+y+%3D+√x+for+x≥0
notice that they both pass though the origin, but after that they never intersect or touch again, and the sine is always above the square root function.
so x ≥ 0
mis-stated the result of the graph
notice that the sin(x) is always below y = √x
so there is no solution to
sin(x) > √x
To solve the inequality sin(x) > √x, we need to find the values of x that satisfy this inequality.
To start, let's square both sides of the inequality:
(sin(x))^2 > (sqrt(x))^2
Simplifying, we get:
sin^2(x) > x
Now, we need to examine the behavior of sin(x) and x separately to determine the values of x that satisfy the inequality.
For sin(x), its values range from -1 to 1. Therefore, sin^2(x) will also range from 0 to 1.
For x, it is a non-negative real number. So, we are only concerned with values of x greater than or equal to 0.
Now, considering the graph of y = sin^2(x), it starts at 0, reaches a maximum at 1, and remains between 0 and 1 as x increases. On the other hand, the graph of y = x starts at the origin and continues to increase indefinitely.
So, we are looking for values of x where sin^2(x) > x. From the behavior of the two functions, we can observe that sin^2(x) > x for values of x between 0 and 1.
Thus, the solution to the inequality sin(x) > √x is: 0 < x < 1.