1)f(x)=4x2−7x+3f′(x)=8x−7f′(x)=0=8x−7x=87f(x) at 87 is 16−1f(x) at -2 and 3 is 33 and 18 respectively∴ absolute max occur at (-2,33)Absolute min occur at (87,16−1)2)f(x)=4x3−8x2+1f′(x)=12x2−16xx=0,34f(0)=1f(34)=27−101f(−1)=−11f(1)=−3absolute max is (0,1) while absolute min is (-1,-11)3)f(x)=x+2x3f′(x)=(x+2)23x2(x+2)−x3f′(x)=0x2−3x−8=0x=4.37,−1.37f(4.37)=13.1f(1.37)=0.76f(−1)=−1f(1)=31Absolute max is (4.37,13.1) and absolute min is (-1,-1)
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