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MIT 18.01 Single Variable Calculus (OCW)
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Finding max and min....... F(x)=2x^3-30x^2+126x+59 explain the steps plz... thanks
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1) Take the derivative first, then set the derivative equal to zero and solve for x. 2) Check the slope on each side of the values found to see if the value is a max or min.
Well, take the derivative of F(x). Then, solve for x. You will find a value, which can be a max or a min. TO know if it' s a max or a min, you have to know the concavity. The derivatives are useful in this case because there is a point which the slope of the derivative is 0. These are the points of interest.
take the 2nd derivative and see if F''(x) >0 or F''(x) <0
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