How do you find the local extremas for #x(x-1)# on [0,1]?

Question

Charles Anctil · Accepted Answer

Finding local extremas involves the first derivative being set equal to 0; then finding out how the derivative acts as you plug in values greater or less than those zeros.

Therefore, we can rewrite #x(x-1) = x^2-x#

#d/dx(x^2-x)=2x-1#

#2x-1=0=>x=1/2#

This lies on the interval #[0,1]#

So, a local extrema is possible, not guaranteed.

However, since the multiplicity of our function is odd, that is:

#(2x-1)^1#

There will be a local extrema at #x=1/2#

Let's plug in #0#.

#2(0)-1=-1=>#negative slope from #[0,1/2]#

Again, since the multiplicity is odd, there will be a positive slope from #[1/2, 2]#

#:.#There is a local minimum at #x=1/2#

Ezekiel Alexander · Answer

undefined To find the local extrema for the function $ f(x) = x(x-1) $ on the interval $[0,1]$, you first need to find its critical points within the interval. Then, you evaluate the function at these critical points as well as at the endpoints of the interval $[0,1]$. The critical points occur where the derivative of the function is either zero or undefined. In this case, the derivative of $ f(x) $ is $ f'(x) = 2x - 1 $.

1. Find the critical points by setting the derivative equal to zero and solving for $ x $:
   $$ 2x - 1 = 0 $$
   $$ x = \frac{1}{2} $$
   
2. Evaluate the function at the critical point and endpoints:
   - $ f(0) = 0(0-1) = 0 $
   - $ f(1) = 1(1-1) = 0 $
   - $ f\left(\frac{1}{2}\right) = \frac{1}{2}\left(\frac{1}{2}-1\right) = -\frac{1}{4} $
   
3. Compare the values obtained. The local extrema occur at the points where the function has maximum or minimum values.
   - Since $ f(0) = 0 $, $ f(1) = 0 $, and $ f\left(\frac{1}{2}\right) = -\frac{1}{4} $, the local maximum is $ x = 0 $ and the local minimum is $ x = \frac{1}{2} $.