18.01 Calculus Jason Starr Fall 2005 Lecture 9. September 29, 2005Homework. Problem Set 2 all of Part I and Part II.Practice Problems. Course Reader: 2B11 2B22 2B44 2B551. Application of the Mean Value Theorem. A realwoorl application of the Mean Value Theorem is error analysis. A device accepts an input signal xand returns an output signal y. If the input signal is always in the range −1/2 ≤ x≤ 1/2 and if the output signal is, 1 y= f(x) = 1 + x+ x2 + x3 , what precision of the input signal xis required to get a precision of ±10−3 for the output signal? If the ideal input signal is x= a, and if the precision is ±h, then the actual input signal is in the range a− h≤ x≤ a+ h. The precision of the output signal is f(x) − f(a). By the Mean Value ||Theorem, f(x) − f(a) = f�(c), x− a for some cbetween aand x. The derivative f�(x) is, f�(x) = −(3x2 + 2x+ 1) . (1 + x+ x2 + x3)2 For −1/2 ≤ x≤ 1/2, this is bounded by, 3(1/2)2 + 2(1/2) + 1 |f�(x)= 7.04.|≤ [1 + (−1/2) + (−1/2)2 + (−1/2)3]2 Thus the Mean Value Theorem gives, f(x) − f(a)= f�(c≤ 7.04x− a≤ 7.04h.|||)||x− a| || Therefore a precision for the input signal of, h= 10−3/7.04 ≈ 10−4 guarantees a precision of 10−3 for the output signal. 2. First derivative test. A function f(x) is increasing, respectively decreasing, if f(a) is less than f(b), resp. greater than f(b), whenever a is less than b. In symbols, f is increasing, respectively decreasing, if f(a) f(b) whenever a1/3, it is negative for −1 1/3. 3. Extremal points. If f(x) ≤ f(a) for all xnear a, then xis a local maximum. If f(x) ≥ f(a) for all xnear a, then xis a local minimum. Because of the First Derivative Test, if f�(a) >0 and f is defined to the right of a, the graph of f rises to the right of a. Thus ais not a local maximum. Similarly, if f�(a) <0 and f is defined to the left of a, the graph of f rises to the left of a. Thus ais not a local maximum. In particular, if f is defined to both the right and left of a, if f�(a) is defined, and if ais a local maximum, then f�(a) equals 0. Similarly, if f is defined to both the right and left of a, if f�(a) is defined, and if ais a local minimum, then f�(a) equals 0. A point awhere f�(a) is defined and equals 0 is a critical point. By the last paragraph, if x= ais a local maximum of f, respectively a local minimum of f, then one of the following holds. (i) The function f(x) is discontinuous at a. (ii) The function f(x) is continuous at a, but f�(a) is not defined. (iii) The point ais a left endpoint of the interval where fis defined, and f�(a) ≤ 0, resp. f�(a) ≥ 0. (iv) The point a is a right endpoint of the interval where f is defined, and f�(a) ≥ 0, rexp. f�(a) ≤ 0. (v) The function f is defined to the left and right of a, and f�(a) equals 0. In other words, ais a critical point of f. 2Example. For the function y= x3 + x− x− 1, the critical points are x= −1 and x= 1/3. By examining where yis increasing and decreasing, x= −1 is a local maximum and x= 1/3 is a local minimum. The plurals of “maximum” and “minimum” are “maxima” and “minima”. Together, local maxima and local minima are called extremal points, or extrema. These are points where f takes on an 18.01 Calculus Jason Starr Fall 2005 extreme value, either positive or negative. A point where f achieves its maximum value among all points where f is defined is a global maximum or absolute maximum. A point where f achieves its minimum value among all points where f is defined is a global minimum or absolute minimum. 4. Concavity and the Second Derivative Test. For a differentiable function f, every “interior” extremal point is a critical point of f. But not every critical point of f is an extremal point. Example. The function f(x) = x3 has a critical point at x= 0. But f(x) is everywhere increasing, thus x= 0 is not an extremal point of f. When is a critical point an extremal point? When is it a local maximum? When is it a local minimum? This is closely related to the concavity of f. A function f(x) is concave up, respectively concave down, if no secant line segment to f(x) crosses below the graph of f, resp. above the graph of f. In symbols, f is concave up, resp. concave down, if (f(c) − f(a))/(c− a) ≤ (f(b) − f(a))/(b− a) whenever ac>b. This precisely says that f� is nondecreeasing resp. f� is nonincreeasing If f� is nondecreeasing resp. nonincreeasing then f is concave up, resp. concave down. Applying the First Derivative Test to determine when f� is increasing, resp. decreasing, gives the Second Derivative Test : If f��(a) >0, then f is concave up near x= a; if f��(a) <0 then f is concave down near x= a. If f is concave up near a critical point, the critical point is a local minimum. If f is concave down near a critical point, the critical point is a local maximum. Combined with the Second Derivative Test, this gives a test for when a critical point is a local maximum or local minimum: If f�(a) equals 0 and f��(a) <0, then x= ais a local maximum. If f�(a) equals 0 and f��(a) >0, then x= ais a local minimum. 2Example. For y = x3 + x− x− 1, the second derivative is y�� =6x+ 2. Since y��(−1) = −4 is negative, the critical point x= −1 is a local maximum. Since y��(1/3) = 4 is positive, x= 1/3 is a local minimum. 5. Inflection points. If f is differentiable, but for every neighborhood of a, f is neither concave up nor concave down on the entire neighborhood, then ais an inflection point. If f��(a) is defined, the Second Derivative Test says that f��(a) must equal 0. Except in pathological cases, an inflection point is a point where f is concave up to one side of f, and concave down to the other side of f. 2Example. For y = x3 + x− x− 1, the second derivative y�� =6x+ 2 is negative for x<−1/3 and is positive for x>1/3. By the Second Derivative Test, y is concave down for x<−1/3 and y is concave up for x>−1/3. Therefore x= −1/3 is an inflection point for y.