SOLUTION: Graph the function f(x)=x^3-2x^2+x by finding its zeros. Use the graph to estimate where the instantaneous rate of change is positive, negative, and zero. Thanks

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Question 1153827: Graph the function f(x)=x^3-2x^2+x by finding its zeros. Use the graph to estimate where the instantaneous rate of change is positive, negative, and zero.
Thanks
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E3-2x%5E2%2Bx
x%28x%5E2-2x%2B1%29
x%28x-1%29%28x-1%29
Zeros at x for -1, 0, and 1.

graph%28400%2C400%2C-3%2C3%2C-3%2C3%2Cx%5E3-2x%5E2%2Bx%29


You could estimate the rate of change information by examining the graph.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Note the factorization and the graph shown in the response from the other tutor are fine; but they have listed the roots incorrectly.

x%5E3-2x%5E2%2Bx+=+x%28x%5E2-2x%2B1%29+=+x%28x-1%29%5E2

There is a single root at x=0 and a double root at x=1; the graph shows that clearly.

graph%28400%2C400%2C-2%2C3%2C-1%2C1%2Cx%5E3-2x%5E2%2Bx%29

For finding where the instantaneous rate of change is positive, negative, or zero, we can first tell that the rate is 0 at exactly x=1 because of the double root. And the graph shows that the rate of change is positive for x > 1 and negative just to the left of x=1.

The graph clearly shows that the rate of change is also zero somewhere between 0 and 1. Since the instructions are to estimate, it looks like the rate is 0 at about x = 1/3. (And calculus would tell us that is exactly right....)

And the rate of change is clearly positive to the left of that point, and negative between that point and x=1.