Finding local extrema and intervals of increase/decrease Example 1 Find the relative extrema of f(x) = increasing and decreasing. ln x x. Also, find where f(x) is STEP 1: Find the domain of the function given. First thing you want to know is which points are allowed to be local extrema. These points have to be part of the domain of the function, so you need to find that first. Looking at the denominator, ln x, we can only input x > 0. The denominator allows everything except x = 0. We conclude that our domain is x > 0 or (0, ). STEP 2: Find the derivative of the given function and simplify. To find the derivative, we will apply the quotient rule. f (x) = x d dx ln x ln x d dx x STEP : Find the critical points. = 1 x x ln x(1) = 1 ln x Critical points are points such that f (x) = 0 or where f (x) is undefined but part of the domain of the original function. First, let s find where f (x) = 0: 0 = 1 ln x 0 = 1 ln x 1 = ln x e = x Next find where f (x) is undefined but part of the domain of the original function:
f 1 ln x (x) = is undefined at x = 0 but x = 0 is not in the domain of f(x) so we do not get anything from here. STEP 4: Make a number line, place critical points on the number line, pick test points between the critical point(s), and plug these test points into the derivative and find sign of the derivative in these intervals. 0 e I will pick a value between 0 and e like say 1, and plug into f (x) f (1) = 1 ln 1 1 2 = 1 > 0. We don t care about the actual number, just the sign of the number. Now, I pick a number greater than e say and plug into my derivative. f () = 1 ln 2.01095 < 0. I will redraw my number line with the signs that I got above. + + + + + - - - - - - - - - 0 e A couple things to note: The number that I picked between 0 and e and greater than e, does not matter. The sign we got from plugging into the derivative does not change if we plug another number in that interval. Also, even though 0 was not a critical point, we place on the number line because our domain restricts us to have x > 0. STEP 5: Interpret the signs we got. On the interval (0, e), f (x) > 0 which means that f(x) is increasing. On the interval (e, ), f (x) < 0 which means that f(x) is decreasing.
Because our function goes from an increasing function to a decreasing function at x = e, we have it that we have a relative maximum at x = e. When your function goes from a decreasing function to an increasing function at some critical point x = a, we have a relative minimum at x = a STEP 6: Write our results. f(x) is increasing on (0, e). f(x) is decreasing on (e, ) f(x) has a relative maximum at e, 1 e Example 2 Find the relative extrema of f(x) = x 16 x STEP 1: Find the domain of the function given. The domain of this function is all real numbers or (, ). Note that we have an odd index on the radical. STEP 2: Find the derivative of the given function and simplify Here we will use the product rule because f(x) is made up of two functions which are being multiplied together. When taking the derivative of 16 x, we will need to use the chain rule. f (x) = x d dx 16 x d + 16 x dx x = x( 1 (16 x) 2 ) + (16 x) 1 = x (16 x) 2 + (16 x) 1 = 4x+48 (16 x) 2 Hence, f (x) = 4x+48 (16 x) 2 STEP : Find the critical points.
First, let s find where f (x) = 0: 4x + 48 0 = (16 x) 2 0 = 4x + 48 12 = x Next find where f (x) is undefined but part of the domain of the original function: f (x) is undefined at x = 16 because this value makes our denominator equal to zero and x = 16 is in the domain of f(x) so this is a critical point. STEP 4: Make a number line, place critical points on the number line, pick test points between the critical point(s), and plug these test points into the derivative and find sign of the derivative in these intervals. 12 16 I will pick a number between (, 12), say 0, and plug into f (x): f (0) = 0 + 48 (16) 2 = 2 2 > 0 I will now pick a number between (12, 16), say 14 into f (x): f (14) = 8 (2) 2 < 0 Finally, I will plug in a number from (16, ), say 17 into f (x): f (17) = 20 ( 1) 2 < 0 I will redraw my number line with the signs that I got above.
+ + + + + - - - - - - - - - 12 16 STEP 5: Interpret the signs we got. On the interval (, 12), f (x) > 0 which means that f(x) is increasing. On the interval (12, ), f (x) < 0 which means that f(x) is decreasing Because our function goes from an increasing function to a decreasing function at x = 12, we have it that we have a relative maximum at x = 12. How about = 16? This point is neither a relative maximum or relative minimum because our derivative did not change sign. We have a vertical tangent at that point. A vertical tangent occurs when you have a point of non-differentiability but you have f (x) does has different signs on either side of that point( like the line above shows) and lim x a+ f (x) = and lim x a f (x) = or vice versa. A picture of a vertical tangent is shown below: Another type of non-differentiability that comes up a lot with these problems is a cusp point. A cusp point occurs when f (x) has the same sign on either side of that point and lim x a f (x) = or lim x a f (x) = A picture of a cusp point is shown below:
STEP 6: Write our results. f(x) is increasing on (, 12). f(x) is decreasing on (12, ) f(x) has a relative maximum at 12, 12 4