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

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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(6294) About Me  (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.
graph%28400%2C400%2C-10%2C10%2C-2%2C20%2C%28x%5E2%29%5E%281%2F3%29%2B5%29
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(3758) About Me  (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 you
  1. Find the values of the function
    • At each of the endpoints of the interval
    • at "bumps" and "pointed-parts" in between the endpoints
  2. Compare 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:
f%28x%29+=+x%5E%282%2F3%29+%2B+5+=+%28root%283%2C+x%29%29%5E2+%2B+5
Now we'll find the values of the functions at the endpoints:
f%28-1%29+=+%28root%283%2C+-1%29%29%5E2+%2B+5
Since root%283%2C+-1%29+=+-1 we get
f%28-1%29+=+%28-1%29%5E2+%2B+5+=+1+%2B+5+=+6

f%288%29+=+%28root%283%2C+8%29%29%5E2+%2B+5
Since root%283%2C+8%29+=+2 we get
f%288%29+=+%282%29%5E2+%2B+5+=+4+%2B+5+=+9

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: 2%2F%283root%283%2C+x%29%29. 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.

f%280%29+=+%28root%283%2C+0%29%29%5E2+%2B+5+=+0+%2B+5+=+5

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).