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# How To Answer THese Calculus Questions? (1-4)?

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- 2 months ago
The critical values are the end points x=-10 and x =1 and where the first derivative is zero (4x^3-16x)=x(4x^2-16) or x=0 and x = plus or minus 2 but note that x=2 is out of the range so we ignore it . Recall when the first derivative =0 we have a possible maximum or minimum...The other critical value sometimes is when the second derivative is set to zero to see if there is an inflection point. 12x^2-16 =0.....once you have all the possible critical x values, just plug each one in to find the corresponding y value.......in this problem, you dont care about the inflection point, so the critical values for x are -2,0 and the two endpoints

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