SOLUTION: Consider the function f(x) =x^5-3*x^3+x-1. Draw a graph of the function on the interval [-1,1} and from the graph determine the point x at which f'(x) is greatest on this interval.

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Question 1049098: Consider the function f(x) =x^5-3*x^3+x-1. Draw a graph of the function on the interval [-1,1} and from the graph determine the point x at which f'(x) is greatest on this interval.
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!

Get some points so you can draw the graph:

   x|y
----------
  -1|0
-0.7|-0.84
-0.5|-1.16
-0.3|-1.22
-0.1|-1.10
   0|-1
 0.2|-0.82
 0.4|-0.78 
 0.5|-0.84
 0.7|-1.16
   1|-2




The curve starts out going downhill to the right, which
means f'(x), the slope of a tangent line, is negative.
The point x at which f'(x) is greatest on this interval 
would have to be a point where the slope is positive.  
That's where the curve is going uphill to the right 
So we look for a point where the curve is the steepest and
going uphill to the right.

I'd guess that the point where the graph is the steepest
going uphill to the right is the point (0,-1). I've drawn a
tangent line there (in green).

So the point x at which f'(x) is greatest on this interval
is where x=0, for the curve is steepest going uphill to the
right there. It's not as steep going uphill to the right at
any other point.

Answer: x=0 for that's the point (0,-1) where the curve is
the steepest going uphill to the right.

Edwin

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