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Find all local maxima and minima and values of x were they occur. f(x)=(x+3)(x-1)^2 (Round to two decimal places)
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f(x)=(x+3)(x-1)(x-1) f(x)-x^2+2x-3(x-1) f(x)=x^3+x^2-5x+3 to find the local max and min you need to take the derivative and set that = 0 f'(x)=3x^2+2x-5=0 (3x+5)(x-1)=0 x=-5/3 x=1 now plug those into your original eq f(x)=(x+3)(x-1)(x-1) testing x=-5/3 f(-5/3)=(-5/3+3)(-5/3-1)^2 f(-5/3)=4/3(-8/3)^2 f(-5/3)=4/3(64/9) f(-5/3)=256/27=9.48 now test x=1 x=1 the function is 0 because (x-1) plug x=1 its 0 so you have a max at (-5/3,9.48) and a min at (1,0)
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