Question

Explain why the function has a zero in the given interval. $$ \begin{array}{lll} \text { Function } & \text { Interval } \\ h(x)=-2 e^{-x / 2} \cos 2 x &{\left[0, \frac{\pi}{2}\right]} \\ \end{array} $$

Short answer

Yes, the function \(h(x)=-2 e^{-x / 2} \cos 2 x\) does have a zero in the given interval [\(0, \frac{\pi}{2}\)]. The zero is at \(x=\frac{\pi}{4}\).

Step-by-step solution

Understand the concept of a zero/root

A root, or zero, of a function is a value \(x\) where the function evaluates to zero. In other words, if there is a value \(x\) in the interval \([0, \frac{\pi}{2}]\) such that when it is substituted into \(h(x)\), the result is zero, then the function has a root or a zero in this interval. We will try to find such a value.

Set the function \(h(x)=-2 e^{-x / 2} \cos 2 x\) equal to zero

The equation to solve is then \(0=-2 e^{-x / 2} \cos 2 x\).

Solving for \(x\)

To solve this equation for \(x\), note that a product of numbers equals zero if and only if at least one of the numbers is zero. Therefore, either \(-2 e^{-x / 2}=0\) or \(\cos 2 x = 0\).\n- The case \(-2 e^{-x / 2}=0\) has no answer because the exponential function is always positive, so the product is never zero.\n- The general solution for \(\cos 2 x = 0\) is \(x=\frac{(2n+1)\pi}{4}\), where \(n\) is an integer.\nBecause the given interval is \([0, \frac{\pi}{2}]\), there is a zero in this interval when \(n = 0\). So, the function \(h(x) = -2 e^{-x / 2} \cos 2 x\) has a zero at \(x=\frac{\pi}{4}\) within the mentioned interval.