1.5 Extrema and the Mean Value Theorem

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1 .5 Extrem nd the Men Vlue Theorem.5. Mximum nd Minimum Vlues Definition.5. (Glol Mximum). Let f : D! R e function with domin D. Then f hs n glol mximum vlue t point c, iff(c) f(x) for ll x D. The vlue f(c) is clled the mximum vlue of f on D. Definition.5. (Glol Minimum). Let f : D! R e function with domin D. Then f hs n glol minimum vlue t point c, iff(c) pple f(x) for ll x D. The vlue f(c) is clled the minimum vlue of f on D. Definition.5.3 (Glol Extreme Vlues). Let f : D! R. The glol mximum nd glol minimum vlues of f re clled the glol extreme vlues of f on D. In this section we will develop tools for finding the extreme vlues of function. The first such tool is the concept of locl mximum or minimum vlues. Definition.5.4 (Locl Mximum). Let f : D! R e function with domin D. Then f hs locl mximum vlue t point c, iff(c) f(x) for ll x close to c. The vlue f(c) is clled locl mximum vlue of f. Definition.5.5 (Locl Minimum). Let f : D! R e function with domin D. Then f hs locl minimum vlue t point c, iff(c) pple f(x) for ll x close to c. The vlue f(c) is clled the locl minimum vlue of f. Definition.5.6 (Locl Extreme Vlues). Let f : D! R e function. The locl mximum nd minimum vlues of f re clled the locl extreme vlues of f on D. Remrk. The term locl is pplied ecuse they re the highest or lowest vlues s compred with vlues nery. Remrk. Not ll functions hve glol mximum vlues or glol minimum vlues. Here re two exmples:. Consider the function f :(0, )! R defined y f(x) =.Bygrphingf, showtht x f hs no glol mximum vlue nd hs no glol minimum vlue.. Consider the function f :(0, )! R defined y f(x) =.Bygrphingf, showthtf x hs no glol mximum vlue nd hs no glol minimum vlue. Theorem.5.7 (Extreme Vlue Theorem). Let f :[, ]! R e continuous function with domin [, ]. Then the following hold:. There is c in [, ] such tht f(c) is the glol mximum vlue.. There is d in [, ] such tht f(d) is the glol minimum vlue. Question. How cn one find the glol extreme vlues of function? See section.5.3. Theorem.5.8. If f hs locl mximum or minimum t c, thenf 0 (c) =0. 5

2 Prolem 30. Consider the function f(x) =x +5x +6. Use Theorem.5.8 s tool for finding locl extreme vlues, if ny exist. Prolem 3. Consider the function f(x) =x 3. Use Theorem.5.8 s tool for finding locl extreme vlues, if ny exist. Prolem 3. Consider the function f(x) = x. Lookingtthegrphofthisfunction, we see tht for c =we hve tht f() is locl minimum vlue. However, Theorem.5.8 will not help us to find it y using the derivtive, ecuse f 0 () is undefined. Remrk. Theorem.5.8 provides tool for finding locl extreme vlues. However, there is no gurntee tht it will lwys work. Definition.5.9. Let f : D! R e function. A point c in D is clled criticl point if one of the following holds:. f 0 (c) =0,or. f 0 (c) is undefined. Theorem.5.0. If function f ttins locl mximum or minimum vlue t point c, thenc is criticl point..5. Finding the Criticl Points Procedure. Given function f, inordertofindthecriticlpointsoff do the following:. Find f 0 (x).. Set f 0 (x) =0ndsolveforx. Iff(x) is defined t ny such solution x, thenyouhve criticlpoint. 3. Determine the vlues of x where f 0 (x) is undefined. If f(x) is defined t ny such vlue of x, thenyouhvecriticlpoint. Prolem 33. Find ll the criticl points of of the following functions. f(x) =x 3 5 (4 x).. f(x) =x 3 +3x 4x f(x) = 3p x. 53

3 .5.3 Finding the Glol Extreme Vlues Procedure (Four Step Method). To find the glol extreme vlues of continuous function f on closed intervl [, ], do the following:. Find the criticl points of f in the open intervl (, ).. Mke tle contining ll the criticl points in the first column nd the vles of f t the corresponding criticl points in the second column. 3. Put the endpoints nd in the ottom of the first column nd the vlues of f t the endpoints in the second column. 4. Locte the mximum vlue in the second column. This is the lrgest vlue of the function f on [, ]. Locte the minimum vlue in the second column. This is the smllest vlue of the function f on [, ]. Prolem 34. Find the glol extreme vlues of the following functions:. f(x) =x 3 3x +on the closed intervl [, 4].. f(x) =(x + x) /3 on the closed intervl [, 3]. Solution. We present solution for item. Let f(x) =x 3 3x +. We will find the extreme vlues of f on the closed intervl [, 4] y following the ove procedure.. Find the criticl points of f in the open intervl (, 4). Since f 0 (x) =3x 6x, we first find the solutions, in the intervl (, 4), totheeqution3x 6x =3x(x ) = 0, which re x =0nd x =. As oth of these roots re in the intervl (, 4), they re criticl points. Since f 0 (x) is defined for ll numers in the intervl (, 4), we conclude tht x =0nd x =re ll of the criticl points.. Mke tle contining ll the criticl points in the first column nd the vles of f t the corresponding criticl points in the second column. Since f(0) = nd f() = 3, we otin the tle Put the endpoints nd 4 in the ottom of the first column nd the vlues of f t the endpoints in the second column. Since f( )= nd f(4) = 7, weotinthe tle Locte the mximum vlue in the second column. Thus, 7 is the lrgest vlue of the function f on [, 4]. Locte the minimum vlue in the second column. Thus, 3 is the smllest vlue of the function f on [, 4]. 54

4 .5.4 The Men Vlue Theorem The Men-Vlue Theorem is key ingredient in the proof of the Fundmentl Theorem of Clculus, which gives connection etween the derivtive nd the re prolem. The Fundmentl Theorem of Clculus nd integrtion re the mjor topics of Clculus II. Theorem.5. (Rolle s Theorem). Let f e differentile on (, ) nd continuous t the endpoints nd. Iff() =f() =0,thenthereistlestonepointc in the intervl (, ) such tht f 0 (c) =0. f (c) = 0 c Figure 3: Illustrtion for Theorem.5. Theorem.5. (Men Vlue Theorem). Let f e differentile on (, ) nd continuous t the endpoints nd. Then there is t lest one point c in the intervl (, ) such tht f 0 (c) = f() f(). f (c) = f() f() f() f() c Figure 4: Illustrtion for Theorem.5. Theorem.5.3. If f 0 (x) =0for ll x in n intervl, then f is constnt on the intervl. Corollry.5.4. If f 0 (x) =g 0 (x) for ll x in n intervl, then f(x) the intervl, where k is some constnt. g(x) =k for ll x on Prolem 35. The following Figure 5 shows the grph of the function f(x) =x 3 + together with the grph of the line y =x +. The men vlue theorem sttes tht there is tngent line to the function f which is prllel to the given line. Find the points,, c such tht f 0 (c) = f() f() is equl to the slope of the line y =x +. 55

5 4 3 y =x + f(x) =x Figure 5: Wht is the slope of the secnt line y =x +?.5.5 Incresing nd Decresing Functions Definition. Let f : I! R e function where I is n intervl. We sy tht f is incresing on I when the following holds: For ll x,x I if x pple x,thenf(x ) pple f(x ). We sy tht f is strictly incresing on I when the following holds: For ll x,x I if x <x,thenf(x ) <f(x ). We sy tht f is decresing on I when the following holds: For ll x,x I if x pple x,thenf(x ) f(x ). We sy tht f is strictly decresing on I when the following holds: For ll x,x I if x <x,thenf(x ) >f(x ). Definition. Let f : I! R e function where I is n intervl. f is monotonic on I if f is either incresing or decresing on I. f is strictly monotonic on I if f is either strictly incresing or strictly decresing on I. Theorem.5.5 (Incresing/Decresing Test). Let f e function which is differentile on the open intervl (c, d): (i) If f 0 (x) 0 for ll x in (c, d), thenf is incresing on the intervl (c, d). (ii) If f 0 (x) > 0 for ll x in (c, d), thenf is strictly incresing on the intervl (c, d). (iii) If f 0 (x) pple 0 for ll x in (c, d), thenf is decresing on the intervl (c, d). (v) If f 0 (x) < 0 for ll x in (c, d), thenf is strictly decresing on the intervl (c, d). Prolem 36. Let f e the function defined y f(x) =x 8x +7. Find the intervls where the function f is strictly incresing nd where it is strictly decresing. Solution. We pply Theorem.5.5. Since f 0 (x) =x 8=(x 4) hs root 4, wefirst focus our ttention of the intervls (, 4) nd (4, ). If x (, 4), thenx<4 nd f 0 (x) =(x 4) < 0. So,f is strictly decresing on (, 4). Ifx (4, ), then4 <xnd f 0 (x) =(x 4) > 0. So,f is strictly incresing on (4, ). Prolem 37. Let T e the temperture function defined y T (x) = 3x +60x +70 were x is the time in seconds. T (x) is the temperture of metl try t time x. Determine when (time intervl) the metl is cooling. 56

6 .5.6 First Derivtive Test We cn use the ove Incresing/Decresing Test to determine if criticl point is locl mximum or locl minimum. Theorem.5.6 (First Derivtive Test). Suppose f is continuous t criticl point c.. f ttins locl mximum vlue t c (see first grph in Figure 6) if () f 0 (x) > 0 ner to the left of c nd () f 0 (x) < 0 ner to the right of c.. f ttins locl minimum vlue t c (see second grph in Figure 6) if () f 0 (x) < 0 ner to the left of c nd () f 0 (x) > 0 ner to the right of c. 3. c is not locl extremum of f if f 0 (x) hs the sme sign ner to the left of c s the sign ner to the right of c. f (c) = 0 f (l) > 0 f (r) < 0 f (l) < 0 f (c) = 0 f (r) > 0 A Locl minimum A Locl mximum Figure 6: First Derivtive Test Prolem 38. Let f e the function defined y f (x) = x3 mximum nd locl minimum vlues of f. Solution. First we find the criticl points of f (x) = x3 f 0 (x) = 3x 6x = 3x(x 3x +. Find ll the locl 3x +. Since ), we hve criticl points 0 nd. We now pply the first derivtive test to ech of these criticl points.. Note tht f 0 (x) = 3x(x ) > 0 ner to the left of 0 nd f 0 (x) = 3x(x ) < 0 ner to the right of 0. So, y Theorem.5.6, f (0) = is locl mximum vlue of f.. Note tht f 0 (x) = 3x(x ) < 0 ner to the left of nd f 0 (x) = 3x(x ) > 0 ner to the right of. So, y Theorem.5.6, f () = 3 is locl minimum vlue of f. 57