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Graphing polynomials Sketch a graph of the | Ch 4.3 - 13E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 13E Chapter 4.3

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 13E

Graphing polynomials ?Sketch a graph of the following polynomials. identify local extrema, inflection points?, and x- ?and y-intercepts when they exist. ? ? ?f(?x) =?? 4 + 4? 3 ? 12?x2

Step-by-Step Solution:
Step 1 of 3

Solution: Step1 Given that f(x)=3*x^4+4*x^3-12*x^2 Differentiate the given equation to find f’(x) we get, f’(x)= 3*4*x^3+4*3*x^2-12*2*x = 12*x^3+12*x^2-24x Again differentiate f’(x) to find f’’(x) we get, f’’(x)=12*3*x^2+12*2*x-24 = 36*x^2+24*x-24 Step2 To get extreme values we have to use f’(x)=0 => 12*x^3+12*x^2-24x=0 =>12x(x^2+x-2)=0 => x(x^2+2x-x-2)=0 => x(x-1)(x+2)=0 Critical points are 0,1 and -2 Step3 To find inflection points we have to use f’’(x)=0 =>36*x^2+24*x-24=0 =>6(6*x^2+4x-4)=0 =>6*x^2+4x-4=0 =>2 (3*x^2+2x-2)=0 =>3*x^2+2x-2=0 => x=(-2± 4 + 24)/6 =>x=-1.21,0.55 Evaluate f’’(x) at the critical points we get, f”(0 )=36*x^2+24*x-24 = 36*(0)^2+24*0-24 =-24<0 has a local maximum at x=0 And f”(1)= 36*x^2+24*x-24 = 36*(1)^2+24*1-24 = 36-0=36>=0 Has a local minimum at x=1 And f”(-2)= 36*x^2+24*x-24 =36*(-2)^2+24*(-2)-24 = 144-48-24=144-72=72>=0 Has a local minimum at x=-2 Step4 The corresponding function values are f(0 )=3*x^4+4*x^3-12*x^2 = 3*(0)^4+4*(0)^3-12*(0)^2 =0+ 0-0=0 , f(1 )= 3*x^4+4*x^3-12*x^2 = 3*(1)^4+4*(1)^3-12*(1)^2 =3+4-12=-5 And f(-2)= 3*x^4+4*x^3-12*x^2 = 3*(-2)^4+4*(-2)^3-12*(-2)^2 =48-32-48=-32 Finally we see that f”(x) changes sign at x=-1.21,0.55 f(-1.21)=-18.36 and f(0.55)=-2.68 Therefore the inflection point is (-1.21,-18.36) and (0.55,-2.68) And the local maximum at (0,0) , Local minimum at (-2,-32) and (1,-5) For x-intercepts, we have to solve the equation f(x)=0 =>3*x^4+4*x^3-12*x^2=0 =>x^2(3*x^2+4*x-12)=0 =>x=-2.77,1.44 Therefore the intercepts are (-2.77,0) and (1.44,0) Step5

Step 2 of 3

Chapter 4.3, Problem 13E is Solved
Step 3 of 3

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Graphing polynomials Sketch a graph of the | Ch 4.3 - 13E