Question 1150690: Show That The Equation x^(2) + 10 Cos(x) = 1 Has At Least One Solution In The Interval [-3,3].
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
At x=0, x^2+10cos(x) = 0+10, which is positive.
At x=pi, x^2+10cox(x) = (pi)^2-10.
Since pi is less than the square root of 10, the function value at x=pi is negative. x=pi is just outside the prescribed range of values for x; but at x=pi the x^2 is increasing rapidly while the 10cos(x) is changing very slowly, so the function value is increasing at x=pi. That means the function value is negative at values of x slightly less than pi -- e.g., at x=3.
Both x^2 and cos(x) are continuous functions, so x^2+10cos(x) is continuous.
Since the function value is 10 at x=0 and negative for values of x close to pi, the function value must be 1 somewhere between 0 and 3.
Note that the function is even; knowing that there is at least one solution on [0,3] means there are at least two solutions on [-3,3].
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