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x^2 + 1 versus 2x - 1. How do we know which is greater?
Please note that x^2 + 1 is a quadratic term that will always be greater. f(x) = x^2 + 1 g(x) = 2x - 1 The 3 cases that we evaluate are x < 0, x = 0, and x > 0. These collectively represent all possibilities for x. For x < 0 then 2x - 1 yields negative values. However, x^2 + 1 has x^2, which is always positive and therefore x^2 + 1 is always positive. Therefore, for x < 0, x^2 + 1 > 2x - 1. Next, when x = 0, then f(x) = x^2 + 1 --> f(0) = (0)(0) + 1 = 1. Thus, x^2 + 1 --> 1. Here, g(x) = 2x - 1 --> g(0) = (2)(0) - 1 = 0 - 1 = -1. Thus, 2x - 1 = -1 when x = 0. Therefore, for x = 0, x^2 + 1 > 2x - 1. Finally, for x > 0, we can look and see that for instance, for x = 1, f(1) = (1)(1) + 1 = 2 and g(1) = (2)(1) -1 = 2 - 1 = 1 so x^2 + 1 > 2x - 1. In fact, the parabolic f(x) = x^2 + 1 will be everywhere above the straight line g(x) = 2x - 1. Here f(x) is growing at the rate: f'(x) = d/dx [x^2 + 1] = 2x + 0 = 2x, which goes up exponentially. On the other hand, g(x) is going up with the slope of 2 so g'(x) = 2. Then, 2x > 2 growth for x > 1. Between 0 < x < 1, f(x) is above g(x) anyways and g(x) = 0 for x = 0.5, while f(x) is always above 0. Therefore, please note that in general for all 3 cases, f(x) = x^2 + 1 is always above g(x) = 2x - 1. Please note the behavior of each of these functions by looking at these lines: f(x) : https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=x%5E2%20%2B%201 g(x) : https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=y+%3D+2x+-+1
What are some of the indeterminate forms that L'Hospital's Rule would help us find the derivative of?
These indeterminate forms, for which L'Hospital's Rule is helpful if if we are finding the limit of f(x) / g(x) where evaluating the limits lead to: (∞/∞, -∞, ∞, 0/0). Other indeterminate forms are like: 1^∞, 0^0, ∞^0, ∞-∞. Here, there is no conceptual way or defined way to evaluate these forms and thus we please have to find the derivatives of f(x) and g(x) and then evaluate the limit of the derivatives: limit of f'(x) / g'(x) and that will be equivalent here.