f(x)=(−34)x4−8x3−(452)x2+105f′(x)=(−34)×4x3−8×3x2−(452)×2x=−3x3−24x2−45x=−3[x2+8x+15]=−3[x2+5x+3x+15]=−3[x(x+5)+3(x2+5)]=−3(x+5)(x+3)nowforcriticalpointsf′(x)=0x=−5andx=−3∴atx=−5isapointofminima∴atx=−3isapointofmaxima.
Find the points of local maxima or minima for the function f(x) =sin2x on [0,π]