Example 1Consider the function f(x) = 1/x on the interval [1, 4]. Note\frac{f(4)  f(1)}{4  1} = \frac{1}{4}. By the mean value theorem, there exists a point c \in [1,4]such that f'(c) = 1/4. We can actually find such a c in this case: c = 2. Example 1Consider the function f(x) = 1/x on the interval [1, 4]. Note\frac{f(4)  f(1)}{4  1} = \frac{1}{4}. By the mean value theorem, there exists a point c \in [1,4]such that f'(c) = 1/4. We can actually find such a c in this case: c = 2. 
Example 2Any two functions with the same derivativeseverywhere will have the same graph except shifted vertically. Example 2Any two functions with the same derivativeseverywhere will have the same graph except shifted vertically. 
Example 3Let f(x) = 2x^3  3x^2  12x + 4. Then f'(x) = 6x^2  6x  12.Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing. When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. There is a local max at x = 1 and a local min at x = 2. Example 3Let f(x) = 2x^3  3x^2  12x + 4. Then f'(x) = 6x^2  6x  12.Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing. When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. There is a local max at x = 1 and a local min at x = 2. 
Example 4Letf(x) = \frac{1}{\sqrt{x^2 + 1}} Then f'(x) = \frac{x}{(x^2 + 1)\sqrt{x^2 + 1}}. Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing.When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. There is a local max at x = 0. Example 4Letf(x) = \frac{1}{\sqrt{x^2 + 1}} Then f'(x) = \frac{x}{(x^2 + 1)\sqrt{x^2 + 1}}. Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing.When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. There is a local max at x = 0. 
Example 5Letf(x) = x^3  3x^2  24x + 5 Then f'(x) = 3x^2  6x  24 Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing.When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. There is a local max at x = 2 and and a local min at x = 2. Example 5Letf(x) = x^3  3x^2  24x + 5 Then f'(x) = 3x^2  6x  24 Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing.When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. There is a local max at x = 2 and and a local min at x = 2. 
Example 6Letg(x) = 3\sqrt[3]{x}(x^2  7) Then g'(x) = 7x^{2/3}(x^2  1) Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing.When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. There is a local max at x = 1 and and a local min at x = 1. Example 6Letg(x) = 3\sqrt[3]{x}(x^2  7) Then g'(x) = 7x^{2/3}(x^2  1) Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing.When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. There is a local max at x = 1 and and a local min at x = 1. 
Example 7Letf(x) = x^3+1 Then f'(x) = 3x^2 Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing.When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. There are no local extrema. Example 7Letf(x) = x^3+1 Then f'(x) = 3x^2 Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing.When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. There are no local extrema. 
Example 7Letf(x) = \frac{3x+1}{x^2+2} Then f'(x) = \frac{3x^2  2x + 6}{(x^2+2)^2} Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing.When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. The local max is at x =\frac{1}{3} \, \sqrt{19}  \frac{1}{3} and the local min is at x = \frac{1}{3} \, \sqrt{19}  \frac{1}{3}. Example 7Letf(x) = \frac{3x+1}{x^2+2} Then f'(x) = \frac{3x^2  2x + 6}{(x^2+2)^2} Here is a plot of y = f'(x), showing where the function is negative and positive:Here is a plot of y = f(x). When f'(x) < 0, f is decreasing.When f'(x) > 0, f is increasing.Also, you can use where the derivative changes signs to find the local extrema. The local max is at x =\frac{1}{3} \, \sqrt{19}  \frac{1}{3} and the local min is at x = \frac{1}{3} \, \sqrt{19}  \frac{1}{3}. 
