Question

In Exercises \(23-28\) , find (a) the local extrema, (b) the intervals on which the function is increasing, and (c) the intervals on which the function is decreasing. $$g(x)=2 x+\cos x$$

Short answer

The function \(g(x) = 2x + \cos (x)\) has no local extrema, is increasing over the interval \((- \infty, \infty)\), and doesn't decrease at any interval.

Step-by-step solution

Obtain the derivative

Differentiate the function \(g(x)\). The derivative of \(g(x) = 2x + \cos (x)\) with respect to \(x\) is given by \(g'(x) = 2 - \sin(x)\).

Find the critical points

Critical points occur in a function when its derivative is either zero or undefined. So, solve \(g'(x) = 0\) for \(x\). \nThis gives: \n\(2 - \sin(x) = 0\). \nHence, \(\sin(x) = 2\). \nHowever, the sine function outputs values between -1 and 1. So, there are no solutions to this equation; implying there are no critical points. Hence, there are no local extrema.

Find the intervals of increasing and decreasing

Given that there are no critical points, meaning there are no turning points or horizontal tangents, the function \(g(x) = 2x + \cos (x)\) will be either decreasing or increasing across its entire domain. Since \(g'(x) = 2 - \sin(x)\) is always greater than zero for all \(x\) as its minimum value is \(2 - 1 = 1\), the function \(g(x) = 2x + \cos (x)\) is increasing over the interval \((- \infty, \infty)\).

Conclusion

Therefore, the function \(g(x) = 2x + \cos (x)\) has no local extrema, always increasing over the interval \((- \infty, \infty)\), and doesn't decrease at any interval.