Given: The function below
[tex]y=\frac{x^2}{x-1}[/tex]
To Determine: If the function as a aximum or a minimum using the first principle
Solution
Let us determine the first derivative of the given function using the first principle
[tex]\begin{gathered} let \\ y=f(x) \end{gathered}[/tex]
So,
[tex]f(x)=\frac{x^2}{x-1}[/tex][tex]\lim_{h\to0}f^{\prime}(x)=\frac{f(x+h)-f(x)}{h}[/tex][tex]\begin{gathered} f(x+h)=\frac{(x+h)^2}{x+h-1} \\ f(x+h)=\frac{x^2+2xh+h^2}{x+h-1} \end{gathered}[/tex][tex]\begin{gathered} f(x+h)-f(x)=\frac{x^2+2xh+h^2}{x+h-1}-\frac{x^2}{x-1} \\ Lcm=(x+h-1)(x-1) \\ f(x+h)-f(x)=\frac{(x-1)(x^2+2xh+h^2)-x^2(x+h-1)}{(x+h-1)(x-1)} \end{gathered}[/tex][tex]\begin{gathered} f(x+h)-f(x)=\frac{x^3+2x^2h+xh^2-x^2-2xh-h^2-x^3-x^2h+x^2}{(x+h-1)(x-1)} \\ f(x+h)-f(x)=\frac{x^3-x^3+2x^2h-x^2h-x^2+x^2+xh^2-2xh-h^2}{(x+h-1)(x-1)} \\ f(x+h)-f(x)=\frac{x^2h+xh^2-2xh+h^2}{(x+h-1)(x-1)} \end{gathered}[/tex][tex]\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{x^{2}h+xh^{2}-2xh+h^{2}}{(x+h-1)(x-1)}\div h \\ \frac{f(x+h)-f(x)}{h}=\frac{x^2h+xh^2-2xh+h^2}{(x+h-1)(x-1)}\times\frac{1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{h(x^2+xh^-2x+h^)}{(x+h-1)(x-1)}\times\frac{1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{x^2+xh-2x+h}{(x+h-1)(x-1)} \end{gathered}[/tex]
So
[tex]\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{x^2-2x}{(x-1)(x-1)}=\frac{x(x-2)}{(x-1)^2}[/tex]
Therefore,
[tex]f^{\prime}(x)=\frac{x(x-2)}{(x-1)^2}[/tex]
Please note that at critical point the first derivative is equal to zero
Therefore
[tex]\begin{gathered} f^{\prime}(x)=0 \\ \frac{x(x-2)}{(x-1)^2}=0 \\ x(x-2)=0 \\ x=0 \\ OR \\ x-2=0 \\ x=2 \end{gathered}[/tex]
At critical point the range of value of x is 0 and 2
Let us test the points around critical points
[tex]\begin{gathered} f^{\prime}(x)=\frac{x(x-2)}{(x-1)^2} \\ f^{\prime}(0)=\frac{0(0-2)}{(0-1)^2} \\ f^{\prime}(0)=\frac{0(-2)}{(-1)^2}=\frac{0}{1}=0 \\ f^{\prime}(2)=\frac{2(2-2)}{(2-1)^2}=\frac{2(0)}{1^2}=\frac{0}{1}=0 \end{gathered}[/tex][tex]\begin{gathered} f(0)=\frac{x^2}{x-1}=\frac{0^2}{0-1}=\frac{0}{-1}=0 \\ f(2)=\frac{2^2}{2-1}=\frac{4}{1}=4 \end{gathered}[/tex]
The function given has both maximum and minimum point
Hence, the maximum point is (0,0)
And the minimum point is (2, 4)
