# SOLUTION: find the absolute maximum and minimum values on the closed interval [-1,8] for the function below. f(x)=x^(2/3)+5 not sure how to due thanks

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: find the absolute maximum and minimum values on the closed interval [-1,8] for the function below. f(x)=x^(2/3)+5 not sure how to due thanks       Log On

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 Click here to see ALL problems on Exponential-and-logarithmic-functions Question 202796: find the absolute maximum and minimum values on the closed interval [-1,8] for the function below. f(x)=x^(2/3)+5 not sure how to due thanks Found 2 solutions by Earlsdon, jsmallt9:Answer by Earlsdon(6287)   (Show Source): You can put this solution on YOUR website!One approach would be to graph the function and get the answers from the graph. You can see from this graph that the minimum in the interval [-1, 8] is f(x) = 0 and the maximumum is f(x) = 9. Answer by jsmallt9(3296)   (Show Source): You can put this solution on YOUR website!Unfortunately Algebra.com's graphing software will not graph this functino. So I will have to describe the solution. (If you have access to a graphing calculator (or some other device that will graph function), use it to see what the the graph of this functino look like. It will help make sense of my explanation.) In general, to find absolute maximum and absolute minimum values youFind the values of the functionAt each of the endpoints of the intervalat "bumps" and "pointed-parts" in between the endpointsCompare all these values and identify the maximum and minimum values. Before we start, I am going to rewrite the function without the fractional exponent because it will help us as we proceed: Now we'll find the values of the functions at the endpoints: Since we get Since we get To find "bumps" and "pointed-parts" of a graph one would usually use Calculus and find where the first derivative is zero (a "bump") or where it does not exist (a "pointed-part"). The first derivative of f(x) is: . Since the numerator is 2 (and cannot become 0), the fraction as a while can never be 0. (Think about it.) On the other hand, if x = 0 then the denominator would end up being 0. But we cannot let this happen. This means the first derivative does not exist when x = 0. (In other words, the graph has a "pointed-part" at x = 0.) So this is one of the points we should check when we are looking for absolute maximum and minimum values of a function. So the absolute maxmum and minimum values must come from these three: f(-1) = 6 f(0) = 5 f(8) = 9 So the absolute maximum value is 9 (when x = 8) and the absolute minimum value is 5 (when x = 0).