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## 4.1 Extreme Values of Functions

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**4.1 Extreme Values of Functions**Absolute extreme values are either maximum or minimum points on a curve. They are sometimes called global extremes. They are also sometimes called absolute extrema. (Extrema is the plural of the Latin extremum.)**4.1 Extreme Values of Functions**• DefinitionAbsolute Extreme Values • Let f be a function with domain D. Then f (c) is the • absolute minimum value on D if and only • if f(x) <f (c) for all x in D. • absolute maximum value on D if and only • if f(x) >f (c) for all x in D.**4.1 Extreme Values of Functions**Extreme values can be in the interior or the end points of a function. No Absolute Maximum Absolute Minimum**4.1 Extreme Values of Functions**Absolute Maximum Absolute Minimum**4.1 Extreme Values of Functions**Absolute Maximum No Minimum**4.1 Extreme Values of Functions**No Maximum No Minimum**4.1 Extreme Values of Functions**Extreme Value Theorem: If f is continuous over a closed interval, [a,b] then f has a maximum and minimum value over that interval. Maximum at interior point, minimum at endpoint Maximum & minimum at interior points Maximum & minimum at endpoints**4.1 Extreme Values of Functions**Local Extreme Values: A local maximum is the maximum value within some open interval. A local minimum is the minimum value within some open interval.**4.1 Extreme Values of Functions**Local extremes are also called relative extremes. Absolute maximum (also local maximum) Local maximum Local minimum Local minimum Absolute minimum (also local minimum)**4.1 Extreme Values of Functions**Notice that local extremes in the interior of the function occur where is zero or is undefined. Absolute maximum (also local maximum) Local maximum Local minimum**4.1 Extreme Values of Functions**Local Extreme Values: If a function f has a local maximum value or a local minimum value at an interior point c of its domain, and if exists at c, then**4.1 Extreme Values of Functions**Critical Point: A point in the domain of a function f at which or does not exist is a critical point of f . Note: Maximum and minimum points in the interior of a function always occur at critical points, but critical points are not always maximum or minimum values.**4.1 Extreme Values of Functions**EXAMPLE 3FINDING ABSOLUTE EXTREMA Find the absolute maximum and minimum values of on the interval . There are no values of x that will make the first derivative equal to zero. The first derivative is undefined at x=0, so (0,0) is a critical point. Because the function is defined over a closed interval, we also must check the endpoints.**4.1 Extreme Values of Functions**At: At: At: To determine if this critical point is actually a maximum or minimum, we try points on either side, without passing other critical points. Since 0<1, this must be at least a local minimum, and possibly a global minimum.**4.1 Extreme Values of Functions**Absolute minimum: Absolute maximum: At: At:**4.1 Extreme Values of Functions**y = x2/3**4.1 Extreme Values of Functions**1 4 Find the derivative of the function, and determine where the derivative is zero or undefined. These are the critical points. For closed intervals, check the end points as well. 2 Find the value of the function at each critical point. 3 Find values or slopes for points between the critical points to determine if the critical points are maximums or minimums. Finding Maximums and Minimums Analytically:**4.1 Extreme Values of Functions**Find the absolute maximum and minimum of the function Find the critical numbers**4.1 Extreme Values of Functions**Find the absolute maximum and minimum of the function Check endpoints and critical numbers The absolute maximum is 2 when x = -2 The absolute minimum is -13 when x = -1**4.1 Extreme Values of Functions**Find the absolute maximum and minimum of the function Find the critical numbers**4.1 Extreme Values of Functions**Find the absolute maximum and minimum of the function Check endpoints and critical numbers The absolute maximum is 3 when x = 0, 3 The absolute minimum is 2 when x = 1**4.1 Extreme Values of Functions**Find the absolute maximum and minimum of the function Find the critical numbers**4.1 Extreme Values of Functions**Find the absolute maximum and minimum of the function The absolute maximum is 1/4 when x = /6, 5/6 The absolute minimum is –2 when x =3/2**4.1 Extreme Values of Functions**Critical points are not always extremes! (not an extreme)**4.1 Extreme Values of Functions**(not an extreme)**4.2 Mean Value Theorem**Mean Value Theorem for Derivatives If f (x) is a differentiable function over [a,b], then at some point between a and b:**4.2 Mean Value Theorem**If f (x) is a differentiable function over [a,b], then at some point between a and b: Mean Value Theorem for Derivatives Differentiable implies that the function is also continuous.**4.2 Mean Value Theorem**If f (x) is a differentiable function over [a,b], then at some point between a and b: Mean Value Theorem for Derivatives Differentiable implies that the function is also continuous. The Mean Value Theorem only applies over a closed interval.**4.2 Mean Value Theorem**If f (x) is a differentiable function over [a,b], then at some point between a and b: The Mean Value Theorem says that at some point in the closed interval, the actual slope equals the average slope. Mean Value Theorem for Derivatives**4.2 Mean Value Theorem**Tangent parallel to chord. Slope of tangent: Slope of chord:**4.2 Mean Value Theorem**(b,0) (a,0) Rolle’s Theorem If f (x) is a differentiable function over [a,b], and if f(a) = f(b) = 0, then there is at least one point c between a and b such that f’(c)=0:**4.2 Mean Value Theorem**Show the function satisfies the hypothesis of the Mean Value Theorem The function is continuous on [0,/3] and differentiable on (0,/3). Since f(0) = 1 and f(/3) = 1/2, the Mean Value Theorem guarantees a point c in the interval (0,/3) for which c = .498**4.2 Mean Value Theorem**(0,1) at x = .498, the slope of the tangent line is equal to the slope of the chord. (/3,1/2)**4.2 Mean Value Theorem**• Definitions Increasing Functions, Decreasing Functions • Let f be a function defined on an interval I and let x1 and x2 • be any two points in I. • f increases on I if x1 < x2 f(x1) < f(x2). • f decreases on I if x1 > x2 f(x1) > f(x2).**4.2 Mean Value Theorem**A function is increasing over an interval if the derivative is always positive. A function is decreasing over an interval if the derivative is always negative. • Corollary Increasing Functions, Decreasing Functions • Let f be continuous on [a,b] and differentiable on (a,b). • If f’ > 0 at each point of (a,b), then f increases on [a,b]. • If f’ < 0 at each point of (a,b), then f decreases on [a,b]. A couple of somewhat obvious definitions:**-**+ + 0 0 f’(x) 2 4 4.2 Mean Value Theorem Find where the function is increasing and decreasing and find the local extrema. x = 2, local maximum x = 4, local minimum**(2,20) local max**(4,16) local min 4.2 Mean Value Theorem**4.2 Mean Value Theorem**Functions with the same derivative differ by a constant. These two functions have the same slope at any value of x.**4.2 Mean Value Theorem**Find the function whose derivative is and whose graph passes through so:**4.2 Mean Value Theorem**Find the function f(x) whose derivative is sin(x) and whose graph passes through (0,2). so: Notice that we had to have initial values to determine the value of C.**4.2 Mean Value Theorem**Antiderivative A function is an antiderivative of a function if for all x in the domain of f. The process of finding an antiderivative is antidifferentiation. The process of finding the original function from the derivative is so important that it has a name: You will hear much more about antiderivatives in the future. This section is just an introduction.**4.2 Mean Value Theorem**Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward. (We let down be positive.) Since acceleration is the derivative of velocity, velocity must be the antiderivative of acceleration.**4.2 Mean Value Theorem**The power rule in reverse: Increase the exponent by one and multiply by the reciprocal of the new exponent. Since velocity is the derivative of position, position must be the antiderivative of velocity.**4.2 Mean Value Theorem**The initial position is zero at time zero.**4.3 Connecting f’ and f’’ with the Graph of f**In the past, one of the important uses of derivatives was as an aid in curve sketching. We usually use a calculator of computer to draw complicated graphs, it is still important to understand the relationships between derivatives and graphs.**4.3 Connecting f’ and f’’ with the Graph of f**local max f’>0 f’<0 local min f’>0 f’<0 no extreme f’>0 f’>0 First Derivative Test for Local Extrema at a critical point c • If f ‘changes sign from positive to • negative at c, then f has a local • maximum at c. 2. If f ‘ changes sign from negative to positive at c, then f has a local minimum at c. • If f ‘ changes does not change sign • at c, then f has no local extrema.**4.3 Connecting f’ and f’’ with the Graph of f**is positive is negative is zero First derivative: Curve is rising. Curve is falling. Possible local maximum or minimum.**4.3 Connecting f’ and f’’ with the Graph of f**concave up concave down • DefinitionConcavity • The graph of a differentiable • function y = f(x) is • concave up on an open interval • I if y’ is increasing on I. (y’’>0) • concave down on an open interval • I if y’ is decreasing on I. (y’’<0)**4.3 Connecting f’ and f’’ with the Graph of f**+ + • Second Derivative Test for Local Extrema at a critical point c • If f’(c) = 0 and f’’(c) < 0, then f has a local maximum at x = c. • If f’(c) = 0 and f’’(c) > 0, then f has a local minimum at x = c.**4.3 Connecting f’ and f’’ with the Graph of f**is positive is negative is zero Second derivative: Curve is concave up. Curve is concave down. Possible inflection point (where concavity changes).