Find an equation of the tangent line to the curve at the given point. y=x3−3x+1,...

Evaluate the derivative using the limit definition, because the easy way is in a later section.
${\displaystyle f\left(x\right)=x^3-3x+1}$, (2,3)
${\displaystyle f‘\left(a\right)=\underset{h\to 0}{lim}\frac{f(a+h)-f\left(a\right)}{h}}$
${\displaystyle =\underset{h\to 0}{lim}\frac{{(a+h)}^3-3(a+h)+1-(a^3-3a+1)}{h}}$
${\displaystyle =\underset{h\to 0}{lim}\frac{a^3+3h{a}^2+3h^{2}a+h^3-3a-3h+1-a^3+3a-1}{h}}$
${\displaystyle =\underset{h\to 0}{lim}\frac{3h{a}^2+3h^{2}a+h^3-3h}{h}}$
${\displaystyle =\underset{h\to 0}{lim}(3a^2+3ha+h^2-3)}$
${\displaystyle =3a^2+0+0-3}$
${\displaystyle =3a^2-3}$
Evaluate the derivative at ${\displaystyle a=2}$
${\displaystyle f‘\left(2\right)=3{\left(2\right)}^2-3=9}$
${\displaystyle f‘\left(x\right)}$ is the slope of the tangent line to ${\displaystyle f\left(x\right)}$ at x, so we have ${\displaystyle m=9}$ at point (2, 3). Plug into the point-slope line equation
${\displaystyle y-y_1=m(x-x_1)}$
${\displaystyle y=m(x-x_1)+y_1}$
${\displaystyle =9(x-2)+3}$
${\displaystyle =9x-18+3}$
${\displaystyle =9x-15}$
Result
${\displaystyle =9x-15}$

The given equation of the curve is ${\displaystyle y=x^3-3x+1}$.
We have to find the tangent line to the curve ${\displaystyle y=x^3-3x+1}$ at point (2, 3).
The point slope form of the tangent line at point ${\displaystyle (x_1,y_1)}$ is given by:
${\displaystyle (y-y_1)=m(x-x_1)}$
Where ${\displaystyle m=\frac{dy}{dx}{\mid }_{(x_1-y_1)}}$ the slope of the tangent line at point ${\displaystyle (x_1,y_1)}$ which is equal to derivative ofthe curve y at point ${\displaystyle (x_1,y_1)}$.
So, slope of the tangent line is given by:
${\displaystyle m=\frac{dy}{dx}{\mid }_{(2,3)}}$
${\displaystyle =\frac{d(x^3-3x+1)}{dx}{\mid }_{(2,3)}}$
${\displaystyle =(3x^2-3){\mid }_{(2,3)}}$
${\displaystyle =3\times 2^2-3}$ {On replacing x with 2}
${\displaystyle =12-3}$
$=9$
Thus, slope ${\displaystyle m=9}$
Now, substitute ${\displaystyle m=9}$ and given point ${\displaystyle (x_1,y_1)=(2,3)}$ in the general form of the line ${\displaystyle (y-y_1)=m(x-x_1)}$.
${\displaystyle =(y-3)=9(x-2)}$
${\displaystyle y-3=9x-18}$
${\displaystyle 9x-y-15=0}$
Hence, required equation of the tangent line to the curve ${\displaystyle y=x^3-3x+1}$ at point (2, 3) is ${\displaystyle 9x-y-15=0}$.