Given f(x)=2x3+15x2−36x+9 Find the absolute maximum and the absolute minimum values of this function.

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Robert Pina · Accepted Answer

Step 1Given function is :f(x)=2x3+15x2−36x+9To find the absolute maximum and the absolute minimum values of this function.Differentiate the function with respect to x ,f′(x)=6x2+30x−36Step 2For maxima or minima f&#039;(x)=0 , therefore6x2+30x−36=06(x2+5x−6)=0(x+6)(x-1)=0x=1, -6Now differentiate f&#039;(x) with respect to x,f&#039;&#039;(x)=12x+30f&#039;&#039;(x) at x=1f″(1)=12×1+30=42&amp;gt;0f&#039;&#039;(1)&amp;gt;0, therefore f(x) has minimum value at x=1Step 3At x=1 , f(x) has minimum value and corresponding minimum value of f(x) is f(1)=2×1+15×1−36×1+9=-10Nowf&#039;&#039;(x) at x=-6f″(−6)=12×(−6)+30=-42&amp;lt;0f&#039;&#039;(-6)&amp;lt;0, therefore f(x) has maximum value at x=-6Step 4At x=-6 , f(x) has maximum value and corresponding maximum value of f(x) is f(−6)=2×(−6)3+15×(−6)2−36×(−6)+9=333Thus, theMaximum value of f(x) is 333Minimum value of f(x) is (-10)

Given f(x) = 2x^{3} + 15x^{2} - 36x + 9

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