Question

determine the maximum and minimum values for the equation : y=2x^3/3+X^2-6x+x

Answer 1

the general procedure you should follow is:
1) you need to find the first differential of y with respect to x,
2) then find what values of x give a zero value for the first differential,
3) then plug each of these x values into the second differential
3.1) if the 2nd differential is negative, then you have a local maximum,
3.2) if its positive, then you have a local minimum,
3.3) if its zero, then you have a point of inflection.
4) plug each x value into the original equation to find the y value at the maximum, minimum, or point of inflection

so, with your equation we get:\[\begin{align}
y&=\frac{2x^3}{3}+x^2-6x+1=\frac{2x^3}{3}+x^2-5x\\
\therefore y'&=\frac{2*3x^2}{3}+2x-6=2x^2+2x-5\\
&\text{following step 2 above, we need to find which values}\\
&\text{of x make y' zero, so:}\\
0&=2x^2+2x-5\\
&\text{solving this the quadratic equation gives:}\\
x&=\frac{-1\pm\sqrt{11}}{2}\\
&\text{we now need to work out the 2nd derivative for step 3 above:}\\
y''&=4x+2\\
&\text{this gives a positive value for } x=\frac{-1+\sqrt{11}}{2}\implies min\\
&\text{and gives a negative value for }x=\frac{-1-\sqrt{11}}{2}\implies max\\
\end{align}\]so now you can follow step 4 above to find the minimum and maximum values of y at each of these x values.

Answer 2

Do we actually NEED to use 2nd derivatives??

Answer 3

The 1st derivative is used to find the points at which the slope is zero.
The 2nd derivative tells you what each of these points represent - i.e. are they maximum, minimum or points of inflection.