91Ó°ÊÓ

First, we need to find the derivative of the given function, \(f(x)=x e^{2x}\). To do this, we will use the product rule, which states that if \(u\) and \(v\) are functions of \(x\), then \((u \cdot v)' = u' \cdot v + u \cdot v'\). Here, we have \(u=x\) and \(v=e^{2x}\). Therefore, \(u'=1\) and \(v'=2 e^{2x}\) (using the chain rule for \(v'\)). Applying the product rule, we get: \(f'(x) = u' \cdot v + u \cdot v'\) \(f'(x) = (1) \cdot e^{2x} + x \cdot 2e^{2x}\) \(f'(x) = e^{2x}(1+2x)\) Now, we are asked to find the values of \(x\) for which the slope of the curve \(y=f(x)\) is 0. In other words, we need to find the values of \(x\) for which \(f'(x) = 0\). So, we set \(f'(x)\) to 0 and solve for \(x\): \(0 = e^{2x}(1+2x)\) Now we have two factors: \(e^{2x}\) and \((1+2x)\). Since exponential functions are never 0, \(e^{2x}\) cannot equal 0. Therefore, we set \((1+2x)=0\): \(0 = 1+2x\) \(2x = -1\) \(x = -\frac{1}{2}\) So, the value of \(x\) for which the slope of the curve \(y=f(x)\) is 0 is \(x=-\frac{1}{2}\).

In part (a), we found that the slope of the curve \(y=f(x)\) is 0 at \(x=-\frac{1}{2}\). The slope of a curve at a certain point represents the steepness of the tangent to that point. When the slope is 0, it means the tangent to the curve is horizontal. This usually happens at points where the function is at a maximum or minimum (local or global) or a point of inflection. In this case, having a slope of 0 at \(x=-\frac{1}{2}\) means that there is either a local maximum or minimum or a point of inflection of the graph of \(f(x) = x e^{2x}\) at that point.