Consider the function f(x)=2x3−6x2−18x+9 on the interval [-2,4]. What is the absolute minimum of f(x) on [-2,4]? What is the absolute maximum of f(x) on [-2,4]?

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Consider the function f(x)=2x3−6x2−18x+9 on the interval [-2,4].   What is the absolute minimum of f(x) on [-2,4]?  What is the absolute maximum of f(x) on [-2,4]?

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Step 1Consider the functionf(x)=2x3−6x2−18x+9, on [-2,4]Step 2The given function is a polynomial so it is continuous everywhere therefore is continuous on [-2,4]Critical points: Critical points of the function are the points where the derivative of the function either zero or does not exist.Compute the derivative of the functionf(x)=2x3−6x2−18x+9f′(x)=6x2−12x−18To find the critical points put f′(x)=0.f′(x)=06x2−12x−18=06(x2−2x−3)=0x2−2x−3=0(x−3)(x+1)=0x=−1,3There are two critical points x=−1 and x=3 and both lies in the interval [-2,4].Step 3Evaluate the function value at the critical points and end points.f(x)=2x3−6x2−18x+9at x=−2f(−2)=2(2)3−6(−2)2−18(−2)+9f(−2)=5at x=−1f(−1)=2(−1)3−6(−1)2−18(−1)+9f(−1)=19at x=3f(3)=2(3)3−6(3)2−18(3)+9f(3)=−45at x=4f(4)=2(4)3−6(4)2−18(4)+9f(4)=−31Absolute maximum value and minimum value is the largest and the smallest function value respectively.Absolute Maximum value =19 at x=−1Absolute Minimum value =−45 at x=−3Step 4Answer:Absolute Maximum value =19Absolute Minimum value =−45

Consider the function f(x)=2x^{3}-6x^{2}-18x+9 on the interval [-2,4]. What is the absolute minimum of f(x) on [-2,4]? What is the absolute maximum of f(x) on [-2,4]?

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