Question

What is the equation of the tangent line of `f(x)=(2x^3 - 1) / (x^2)` at `x=1`?

Answer 1

`y=4x-3`

Explanation:

`"the slope of the tangent line is given by "f'(x)`

`"differentiate using the "color(blue)"quotient rule"`

`"Given "f(x)=(g(x))/(h(x))" then"`

`f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"`

`g(x)=2x^3-1rArrg'(x)=6x^2`

`h(x)=x^2rArrh'(x)=2x`

`rArrf'(x)=(6x^4-2x(2x^3-1))/(x^4)`

`color(white)(rArrf'(x))=(2x^4+2x)/x^4`

`rArrf'(1)=4/1=4larrcolor(blue)"slope of tangent"`

`rArrf(1)=1/1=1rArr(1,1)larrcolor(blue)"point on tangent"`

`rArry-1=4(x-1)larrcolor(blue)"point-slope form"`

`rArry=4x-3larrcolor(red)"equation of tangent"`

Answer 2

The tangent line should be `y=4x-3`

Explanation:

The first step is do determine the slope of the curve at `x=1`. We will do this by taking the derivative of the function.

`f(x)=(2x^3-1)/x^2=2x-1/x^2`

`f(x)=2x-x^(-2)`

`f'(x)=2+2x^-3`

Now, evaluate the derivative to find the slope of the curve at `x=1`:

`f'(1)=2+2(1^(-3))= 2+2`

`color(blue)(f'(1)=4`

Now that we know the slope, we need to know the intercept. We'll figure this out by solving for f(x) and backing out the intercept.

`f(1)=(2(1)^3-1)/((1)^3)= (2*1-1)/1=2-1=1`

`f(1)=color(blue)(4)x+b`

`1=color(blue)(4)(1)+b=4+b`

`color(purple)(b=-3`

Now we have our equation:

`y=color(blue)(4)xcolor(purple)(-3)`