Find the absolute maximum and absolute minimum values of f on the given interval. f(x)=2x3−6x2−18x+9,[−2,4] absolute minimum value absolute maximum value

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Find the absolute maximum and absolute minimum values of f on the given interval.  f(x)=2x3−6x2−18x+9,[−2,4]  absolute minimum value  absolute maximum value

Lynne Trussell · Accepted Answer

Step 1Finding the derivative of the given functionf′(x)=ddx(2x3−6x2−18x+9)=6x2−12x−18To find the critical value we have,f&#039;(x)=06x2−12x−18=0x2−2x−3=0x2−3x+x−3=0x(x-3)+1(x-3)=0(x+1)(x-3)=0x=-1 or x=3So the values between the interval [-2,-1,3,4]Step 2Absolute maximum and minimumf&#039;(-2)=-16-24+36+9=45-40=5f&#039;(-1)=-2-6+18+9=19f&#039;(3)=54-54-54+9=-45f&#039;(4)=128-96-72+9=-31From the above equations, we see that the function has a maximum value at -1 and a minimum at 3.So,Absolute Maximum at x=-1 is 19Absolute Minimum at x=3 is -45

Papilys3q · Accepted Answer

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 function f(x)=2x3−6x2−18x+9f′(x)=6x2−12x−18To find the critical points put f&#039;(x)=0.f&#039;(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= 19 Absolute Minimum value= -45

Find the absolute maximum and absolute minimum values of f

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