Question

solve block on mathcad

I 1. Given the function f(x) = x3 + 10x2 - 11x - 125. a. Graph f(x), showing all roots between -15 SX 55. b. Find the location and value of the local minimum for-55x55 using a solve block.

Answer 1

Solution & alven, f(a) = x3 +10% - 118-125 itt a polynomial funchion Graph, roots on. [-15,5 a at x20= f(2) = ofloxo - 1180- 125 F-125 at 82-15 > fo=(-15} thox-is]) - 11X(-15)-125 - 3375 + 2250 + 165-125 IFCIS) = -1085 +(10) = 610² 10x610) 711x410) 125 =-1000 1000 +10-125 iff&lo) = – 15 f(-5)=653 + 10x (-557_11*(-5)-125 3-125+ 250+55 -125 = f(-5)=55 : +(-10) = -15 co2 Therefore, Jas f(x) is f(-5) = 55 > 0 continuous in [15,5] there exists at least one root between - 10 and 5..

and f(-5)= 55 >.01 fco)=-125 Lo 2. There exists at least one root between x=-5 and 20 "Letus randomly assume some x=3,5 fE8.5)=(-3.5 +10 6-3.5% 11x6-3.5) -125 = -42.875 + 122.5738.5 - 125 16-3-5) =-6.875 <0 and f(-5)=55>0 Root in between - 5 &-3.5 Leto ass ume x=-3.657 3 f(-3.657) = (-3-657) + 10 (-3.657) 1x (-3-657) 125 =-48.9074† 133-7364 +40-227-125 sfc-3-657) = 0.056 x=-3-657 is one of the root of foxy.

we know that if *= a is a root of tow hen Ge-a) will be a factor for fca). re f(x) = (2-63.65 +] { fc) = (x+3.657)$ x+6-3432 - 34.1967 . To find roots feas=0 x+2.657 art 6-3434-34-196) =0 5 X+3.657 d=-3.657 x+6.3438-34.19620 x--6-3431.516-2433 - Чx tx (-34-196): 2x I 32-6-343 I 13.3047 2 x=-6.343 +13.3047 2 3x=3.48085 x=-6-348-13.3047 2 x = -9.82385 Roots of few are: > Answer X=-9082385, 1 X = -3.657 2z=3.48085

Now, To draw f(x) tau= 23+10x²_112-125 f'ca)= 3x² + 200 -11 f'(x)=0 30²7 202-1120 x = -20120_4x3x(+11) 2x3 ax=-20123.065 = Critical points are > x=-701775 and x = 0.5108 7.1775 interval (-15 - 28235) 71775 f'(x) =.1 Let x=-10 € (-15, pares) f(-10) = 3(-103+206-10)-11 =Qx100)+ 20640) - = 300-200- 11 89.>o on interval (-15, 8258) if be) -7.1775 - is increasing fl(a) = 6x+20 an interet 66 396) for inflection points fox)=0 Gat 20 - x= -20 -. (=-3.333 inflection point = -3.333.

f'(a) in l-is, -3,333) let f"(-10)= 6(-10) 150 Gotoo -4o co on -Concare down in (45,-2.330) interval (3.333,5) let azo & Co)= 6x0+20 22000 Cone Concave up in (-3.333,5) at = -7.1775 +(x)= (-7-1775) + 10 (-7-17753-11x4 -7-1775) 125 to f(-7-1715)=-36907597 +515. 165 +78-9525 If(-7-1715) - 125 99.3578 +60:5108) = (0.5108) ²+10 (0.5108,3 - 1X0-5108 -125 If (0-51087 -12 72876 how on interval 67.1775, 0.5108) let x20 flex)=? f'(o)= 3x0 + 2000 112-11 <0

..l-fex) is decreasing on (-74795, 0.5108), on 6.5108, 5): I ca= ? Say +62)=2x2 4 + 2012 - 11 82 12+ yo-11 52-1241 >O on (0.5102,5) ffeses is increasing Now, by combining all those as From on E15.-7.1775] ! increasing & Concare down on C-71775,-3-333] decreasing & Concare down on [- 3.333, 0.5108], decreasing & Concare up. Co-5108,5), increasing & Concave up. and f(-15) = -1085; f(-9-82385)=0 +(-7.1775)=99.3578; f(-3.657)_0; 760-5108) = -127.876 463.48085920) fez

f(s)=(5)²+10x5_1185-125 = 125 +250 - 55-125 if(5)= 195. graphy fow_99.3578 195 > Ha=195. X-9.82395 $20.5108 f -15 it lot + x=7.1775 te=3.48085 (How 725 l=-3.657 Het-1272876) Local minimum on [95] -1085 1 fon)= -1085 Concare down on * K Concave cup: Dans:- from all the above calculation and graph we can clearly observe that local minimumon [-55] is at x=0.5108 and that value į-127.876.