I'm finding the extrama min and max of this problem:G(x)= -x^3+x^2+x-5
I differentiated to -3x^2+2x+1
factored that -(3x+1) (x-1)
getting x to equal 1 or -1/3
by plugging those into the equation to see what both would result as the min being (-1/3,-140/27) and a max of (1,-4). Is this correct?
I differentiated to -3x^2+2x+1
factored that -(3x+1) (x-1)
getting x to equal 1 or -1/3
by plugging those into the equation to see what both would result as the min being (-1/3,-140/27) and a max of (1,-4). Is this correct?
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Yes, it's right. Use your calculator to check the answer next time. Just plug in the original equation, and see if the local minimum and maximum are equal to about the same coordinates that you can visualize by looking at the curve. That's what I did to check yours.
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you are probably correct, but you should check what is happening on the intervals
(- infinity, -1/3) (- 1/3 , 1) and (1, infinity)
interval (- infinity, -1/3)
the two factors are neg, remember the factor out -1
f'(x) < 0
the function is decreasing to x = - 1/3
interval (-1/3, 1)
the function has to increase to go from (-140/27) to -4 but perhaps it is worth verifying
choose a point in the interval, x = 0 is good
f'(x) = 1
f'(x) > 0 on the interval (-1/3, 1)
interval (1, infinity)
the two factors are positive, remember the factored out -1
f'(x) < 0
the function is decreasing from (1, infinity)
Yes, you are correct
(- infinity, -1/3) (- 1/3 , 1) and (1, infinity)
interval (- infinity, -1/3)
the two factors are neg, remember the factor out -1
f'(x) < 0
the function is decreasing to x = - 1/3
interval (-1/3, 1)
the function has to increase to go from (-140/27) to -4 but perhaps it is worth verifying
choose a point in the interval, x = 0 is good
f'(x) = 1
f'(x) > 0 on the interval (-1/3, 1)
interval (1, infinity)
the two factors are positive, remember the factored out -1
f'(x) < 0
the function is decreasing from (1, infinity)
Yes, you are correct
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You can check as follows:
Type of point dy/dx d2y/dx^2
Maximum 0 negative
Minimum 0 positive
Point of inflection Any value 0
Type of point dy/dx d2y/dx^2
Maximum 0 negative
Minimum 0 positive
Point of inflection Any value 0