Math21wa5ch - =expx = fx exx2 x 1 x=exp'x expx x= ex2x 1 ex(x2 x-1 x= ex(x2 3x e x=exx2 3xex divide both sides by ← ex =x2 3x =xx 3 factor out

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Unformatted text preview: =expx = + - fx exx2 x 1 x=exp'x+expx x= ex2x+1+ex(x2+x-1) x= ex(x2+3x) e x=exx2+3xex divide both sides by ← ex =x2+3x =xx+3 factor out ← x =0, -3-3- =- . f' 1 0 7357 = . f'1 8 1548 =exp'x+expx = ex2x+1+ex(x2+x-1) = ex(x2+3x) =exx2+3xex divide both sides by ← ex x2+3x xx+3 factor out ← x 0, -3 f''x=exp''x+exp'(x)+[exp'x+expx] px=x3; = p'x 3x2 ; = p''x 6x f'x=3x2ex+x3ex 0ex=3x2ex+x3exex 0=3x2+x3 0=x2(x+3) x=0, -3 f''x=6xex+3x2ex+[3x2ex+x3ex] 0=ex(x3+6x2+6x) 0ex=x3+6x2+6xex 0=x(x2+6x+6) 0=xx+1.2679(x+4.7321) x=0, -1.2679, -4.7321 Math 21 Written Assignment 5 A) A function has a horizontal tangent line at all points where the first derivative of the function is equal to zero. To differentiate the function f(x), we must use the product rule. We then set the derivative equal to zero and solve for x. The graph of f has horizontal tangent lines at x equals zero and negative three To find if these point are local minimums or local maximums, we must evaluate the derivative of the function at a point greater than zero, a point between zero and negative three, and a point less...
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This note was uploaded on 11/16/2009 for the course MATH 021 taught by Professor Muralee during the Spring '08 term at Lehigh University .

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Math21wa5ch - =expx = fx exx2 x 1 x=exp'x expx x= ex2x 1 ex(x2 x-1 x= ex(x2 3x e x=exx2 3xex divide both sides by ← ex =x2 3x =xx 3 factor out

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