# calculate smallest and highest value of the function

• Jun 28th 2010, 11:09 AM
Garas
calculate smallest and highest value of the function
calculate smallest and highest value of the function \$\displaystyle y=x^3-2x*|x-2|\$ on segment [0,3]
• Jun 28th 2010, 10:24 PM
ecnanif
You check the values of the function at the endpoints, that is at the points \$\displaystyle x=0 \$ & \$\displaystyle x=3 \$.
Now, since your function is not differentiable at \$\displaystyle x=2\$, you cannot use the derivative at \$\displaystyle x=2\$.

So if you limit yourself to \$\displaystyle (0,2)\$ and \$\displaystyle (2,3)\$ when performing you differentiate and look for max/min, and check the function value at \$\displaystyle x=2\$.

Now you compare all the values you have. That is, you have your endpoint values, your value at \$\displaystyle x=2\$, and possibly some values at local max/min which you may have found during the differentiation.
• Jun 28th 2010, 10:43 PM
CaptainBlack
Quote:

Originally Posted by ecnanif
You check the values of the function at the endpoints, that is at the points \$\displaystyle x=0 \$ & \$\displaystyle x=3 \$.
Now, since your function is not differentiable at \$\displaystyle x=2\$, you cannot use the derivative at \$\displaystyle x=2\$.

So if you limit yourself to \$\displaystyle (0,2)\$ and \$\displaystyle (2,3)\$ when performing you differentiate and look for max/min, and check the function value at \$\displaystyle x=2\$.

Now you compare all the values you have. That is, you have your endpoint values, your value at \$\displaystyle x=2\$, and possibly some values at local max/min which you may have found during the differentiation.

Differentiating is not an option, this is not a calculus problem.

CB
• Jun 29th 2010, 06:53 AM
HallsofIvy
Quote:

Originally Posted by Garas
calculate smallest and highest value of the function \$\displaystyle y=x^3-2x*|x-2|\$ on segment [0,3]

If x< 2, x-2< 0 so |x-2|= -(x-2)= 2- x. Then \$\displaystyle y= x^3- 2x(2- x)= x^3+ 2x^2- 4x= x(x^2+ 2x- 4)\$. You can complete the square to determine where \$\displaystyle x^2+ 2x- 4\$ has max or min.

If x> 2, x-2> 0 so |x-2|= x- 2. Then \$\displaystyle y= x^3-2x(x-2)= x^3- 2x^2+ 4x= x(x^2- 2x+ 4)\$. Again, try completing the square in \$\displaystyle x^2- 2x+ 4\$.