4.2: What Derivatives Tell Us

Transcription

1 4.2: What Derivatives Tell Us Problem Fill in the following blanks with the correct choice of the words from this list: Increasing, decreasing, positive, negative, concave up, concave down (a) If you know f 00 (x) > 0, then you know f 0 (x) is and f(x) is. increasing, concave up (b) If you know g 0 (x) < 0 and decreasing, then you know g(x) is and. decreasing, concave down (c) If you know h(x) is positive, increasing, and concave down, then you know h 0 (x) is and and that h 00 (x) is. positive, decreasing, negative Problem 2 Sketch a graph of a function that is continuous on (, ) that has the following properties. (a) Function f does not have a local maximum or minimum. f contains a point where f 0 (x) =0 (b) g 0 (x) < 0 on (, ); g 0 (x) > 0 on (, 2); g 0 (x) < 0 on (2, ).

2 Problem Let f(x) = on the interval [ 2, 2]. Find the following for f: +x2 (a) f 0 and f 00 f 0 (x) = ( + x2 )(0) (2x) ( + x 2 ) 2 2x = ( + x 2 ) 2 (b) Critical points f 00 (x) = ( + x2 ) 2 ( 2) ( 2x)(2)( + x 2 )(2x) ( + x 2 ) 4 = 2( + x2 )+8x 2 ( + x 2 ) = 6x2 2 ( + x 2 ) Since +x 2 > 0 for all x, f is di erentiable over all real numbers. Thus all critical points of f occur when f 0 (x) =0and x 2 ( 2, 2). But a fraction equals 0 if and only if its numerator equals 0. So f 0 (x) =0 =) 2x =0 =) x =0 Hence, the only critical point is x =0. (c) Intervals where f is increasing and decreasing 2

3 We can arrive at the answer two ways: First, we can look at f 0 2x (x) = ( + x 2 and see that the denominator will be positive for all x. ) 2 The sign of f 0 depends on the numerator. If x<0, thef 0 > 0. If x>0,f 0 < 0. Therefore f is increasing on ( 2, 0) and decreasing on (0, 2). On the other hand, we can perform computations: Because f 0 doesnt change sign in the interval ( 2, 0), we can use a test point in this interval to determine the sign of f 0 on the interval. Let us chose x = and, similarly on (0, 2), choose x =. So we have the following picture: f 0 ( ) = 2 4 > 0 f 0 () = 2 4 < 0 (+) (-) f f 0 ( ) = 2 f 0 () = 2 Thus, f 0 is positive on the interval ( ( 2, 0) and decreasing on (0, 2). 2, 0) and negative on (0, 2), and therefore f is increasing on (d) Local extrema (and check your answers with both the first and second derivative tests) At x =0, f 0 changes sign from positive to negative. Thus f goes from increasing to decreasing, and therefore by the first derivative test x =0is a local maximum of f. For the second derivative test, we have that f 00 (0) = 6(0)2 2 ( ) = 2 = 2 < 0 and thus we again conclude that x =0is a local maximum of f. (e) Intervals of concavity To find points were f 00 may change signs, we need to find where f 00 (x) =0or f 00 (x) does not exist. f 00 (x) exists everywhere in our domain so we are only looking for where f 00 (x) =0. So we solve 6x 2 2 ( + x 2 ) =0 6x 2 2=0 x 2 = 2 6 x = ± p

4 We can determine intervals of concavity two ways: First, since the denominator of f 00 is always positive, f 00 has the same sign as the numerator. The factors of the numerator are x p x + p. When x> p, both factors are positive and f 00 > 0. Whenx< p, both factors are negative and f 00 > 0. When negative and the other is positive so f 00 < 0. Thus, f is concave down on up on 2, p and p, 2 p <x< p, one factor is p, p and concave The other way we go about this is using test points: Picking test points of f 00 ( ) = = 4 8 > 0 f 00 (0) = 0 2 = 2 < 0 f 00 () = = 4 8 > 0, 0, and, we have that So we have the following picture: (+) (-) (+) f 00 2 p p 2 f 00 ( ) = 2 f 00 (0) = 2 f 00 () = 2 Thus, f is concave down on 2, p and p, 2 p, p and concave up on (f) Inflection points By the results in part (e) and since f p = 4 f p = 4 the inflection points are: p,, p, 4 4 4

5 (g) Absolute extrema We need to check the critical points of f and the endpoints of the interval: f( 2) = 5 f(0) = f(2) = 5 So the absolute maximum value of f on [ 2, 2] is f(0) =. Both f( 2) = 5 and f(2) = 5 absolute minimum values of f on [ 2, 2]. The graph of the function looks like: are the 2 p p 2 Problem 4 (a) The (entire) graph of a function f is shown in the figure below. 5

6 (i) Find the x-coordinates of all critical points of f (or write NONE). f has a critical point at x =0since f 0 (0) = 0 f has a critical point at x =since f 0 () DNE. f has a critical point at x =since f 0 () DNE. (ii) Find the x-coordinates of all local minima of f (or write NONE). By first derivative test there is a local minimum at x = (iii) Find all values of x at which f attains its global minimum (or write NONE). f() = 0 and lim f(x) = 2 (but f( 2) DNE) imply f has no global minimum x! + (iv) Find the interval (or intervals) on which the derivative of f is increasing. f 0 is increasing on (0,): f is concave up on this interval (b) The figure below shows the graphs of f, f 0, and f 00.Whichcurveiswhich? Graph A is f, Graph B is f 0 and Graph C is f 00. Problem 5 Given the graph of h 0, sketch a graph of h and h 00 on the same set of axis. 6

7 To sketch h 00 from h 0, we need to think back to section.2. It might be easier for you to think of h 0 = g and 7

8 h 00 = g 0. We re really just sketching a graph of the derivative from the graph of its function. We need to recall when the graph of: h 0 = g is increasing, h 00 = g 0 is positive h 0 = g is decreasing, h 00 = g 0 is negative h 0 = g has a local maxima or minima, h 00 = g 0 =0 h 0 = g has an inflection point, h 00 = g 0 has a local maxima or minima h 0 = g is concave up, h 00 = g 0 is increasing h 0 = g is concave down, h 00 = g 0 is decreasing =) The graph of h 00 = g 0 is/has: Positive on the intervals (, 0) and (4, ) Negative on the interval (, ) and (0, 4) Local maximum at x =.5, local minimum at x =2 To sketch h from h 0, we are using the same information, but in reverse. The graph of h is/has: Positive slope on the intervals (, 2) and (5.5, ) Negative slope on the interval (2, 5.5) Local maximum at x =2and local minimum at x =5.5 Inflection points at x =, 0, 4 Concave down on the intervals (, ) and (0, 4) Concave up on the intervals (, 0) and (4, ) Problem 6 Give an example or sketch of a function that is continuous on ( properties. If such a function does not exist, explain why., ) and satisfies given (a) A function f is concave up and negative everywhere. This is not possible. If a function is always concave up then at some point, the function must cross the x-axis and become positive. See the three example figures below. 8

9 (b) A function f is decreasing and concave up everywhere. This is possible. See figure B above. (c) A function s has exactly local extrema and four inflection points. This is possible. See the figure below. The graph of s has inflection points at a, b, c, d and local extrema at e, f, g. (d) A function f has exactly 2 zeros and one local extrema. This is possible. See the figures below for examples. Problem 7 Consider the parabola f(x) =ax 2 + bx + c where a, b, c are constants. For what values of a, b, c is f concave up? For what values of a, b, c is f concave down? f 0 (x) =2ax + b f 00 (x) =2a =) the sign of a will determine is f is concave up or down. When a<0, f is concave down. When a>0, f is concave up. Note: If a =0, f has no concavity because it is a linear function. Problem 8 Locate the critical points and use the second derivative test to determine whether they correspond to local maxima or local minima. f(x) =(x + c) 4 where c is a positive constant 9

10 f 0 (x) = 4(x + c),f 00 (x) = 2(x + c) 2 To find the critical points: f 0 (x) = 0 = 4(x + c).thisoccurswhenx = Using the second derivative test: f 00 ( c) = 2( c + c) 2 =0. The second derivative test was inconclusive so we need to use the first derivative test. Making a sign chart we see that if we plug something a little less than c into f 0, we ll get a negative number. If we plug something a little greater than c into f 0, we ll get a positive number. c The derivative of f, f 0, does not change sign on the intervals (, c) and on ( c, ), therefore, we have the above chart. From the chart, we can conlude that f has a local minimum at x = c, and that f also has a global minimum there. We could also have used our understanding of the family of functions g(x) =x 4 to reason x = c would be a minimum. 0