Question

Using first principles to find derivatives grade 12 calculus help image attached much appreciated

Answer 1

Given: The function below

`y=(x^2)/(x-1)`

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

`\begin{gathered} let \\ y=f(x) \end{gathered}`

So,

`f(x)=(x^2)/(x-1)`
`\lim_(h\to0)f^(\prime)(x)=(f(x+h)-f(x))/(h)`
`\begin{gathered} f(x+h)=((x+h)^2)/(x+h-1) \\ f(x+h)=(x^2+2xh+h^2)/(x+h-1) \end{gathered}`
`\begin{gathered} f(x+h)-f(x)=(x^2+2xh+h^2)/(x+h-1)-(x^2)/(x-1) \\ Lcm=(x+h-1)(x-1) \\ f(x+h)-f(x)=((x-1)(x^2+2xh+h^2)-x^2(x+h-1))/((x+h-1)(x-1)) \end{gathered}`
`\begin{gathered} f(x+h)-f(x)=(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)=(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)=(x^2h+xh^2-2xh+h^2)/((x+h-1)(x-1)) \end{gathered}`
`\begin{gathered} (f(x+h)-f(x))/(h)=(x^(2)h+xh^(2)-2xh+h^(2))/((x+h-1)(x-1))/ h \\ (f(x+h)-f(x))/(h)=(x^2h+xh^2-2xh+h^2)/((x+h-1)(x-1))*(1)/(h) \\ (f(x+h)-f(x))/(h)=(h(x^2+xh^-2x+h^))/((x+h-1)(x-1))*(1)/(h) \\ (f(x+h)-f(x))/(h)=(x^2+xh-2x+h)/((x+h-1)(x-1)) \end{gathered}`

So

`\lim_(h\to0)(f(x+h)-f(x))/(h)=(x^2-2x)/((x-1)(x-1))=(x(x-2))/((x-1)^2)`

Therefore,

`f^(\prime)(x)=(x(x-2))/((x-1)^2)`

Please note that at critical point the first derivative is equal to zero

Therefore

`\begin{gathered} f^(\prime)(x)=0 \\ (x(x-2))/((x-1)^2)=0 \\ x(x-2)=0 \\ x=0 \\ OR \\ x-2=0 \\ x=2 \end{gathered}`

At critical point the range of value of x is 0 and 2

Let us test the points around critical points

`\begin{gathered} f^(\prime)(x)=(x(x-2))/((x-1)^2) \\ f^(\prime)(0)=(0(0-2))/((0-1)^2) \\ f^(\prime)(0)=(0(-2))/((-1)^2)=(0)/(1)=0 \\ f^(\prime)(2)=(2(2-2))/((2-1)^2)=(2(0))/(1^2)=(0)/(1)=0 \end{gathered}`
`\begin{gathered} f(0)=(x^2)/(x-1)=(0^2)/(0-1)=(0)/(-1)=0 \\ f(2)=(2^2)/(2-1)=(4)/(1)=4 \end{gathered}`

The function given has both maximum and minimum point

Hence, the maximum point is (0,0)

And the minimum point is (2, 4)