9-14. Graphing polynomials Sketch a graph of the following polynomials. Identify local extrema, inflection points, and x- and y-intercepts when they exist.

\(f(x)=x^{3}-33 x^{2}+216 x-2\)

Solution: Step1 Given that f(x)=x^3-33*x^2+216x-2 Differentiate the given equation to find f’(x) we get, f’(x)= 3*x^2-33*2*x+216 = 3*x^2-66x+216 Again differentiate f’(x) to find f’’(x) we get, f’’(x)=3*x^2-66x+216 = 3*2*x-66 =6x-66 Step2 To get extreme values we have to use f’(x)=0 => 3*x^2-66x+216=0 =>3(x^2-22x+72)=0 => x^2-22x+72=0 => x^2-4x-18x+72=0 =>x(x-4)-18(x-4)=0 =>(x-18)(x-4)=0 => x=18,4 Critical points are 4,18 Step3 To find inflection points we have to use f’’(x)=0 =>6x-66=0 =>6(x-11)=0 =>x-11=0 =>x=11 Evaluate f’’(x) at the critical points we get, f”(4 )=6x-66 = 6*4-66=24-66 =-42<0 has a local maximum at x=4 And f”(18)= 6x-66 = 6*18-66=108-66 =42>0 Has a local minimum at x=18 Step4 The corresponding function values are f(4 )=x^3-33*x^2+216x-2 = (4)^3-33*(4)^2+216*4-2 =64-33*16+864-2 =64-528+864-2 =928-530=398 and f(18 )= x^3-33*x^2+216x-2 = (18)^3-33*(18)^2+216*18-2 =5832-33*324+3888-2 = 5832-10692+3888-2 =-974 Finally we see that f”(x) changes sign at x=11 f(11)=-288 Therefore the inflection point is (11,-288) And the local maximum at (4,398) , Local minimum at (18,-974) For x-intercepts, we have to solve the equation f(x)=0 =>x^3-33*x^2+216x-2=0 Note :- Difficult to solve for x. I am getting complex value for x. Step5