lilyta79jd

2021-11-19

Find an equation of the tangent line to the curve at the given point.

Antum1978

Beginner2021-11-20Added 15 answers

Evaluate the derivative using the limit definition, because the easy way is in a later section.

$f\left(x\right)={x}^{3}-3x+1$ , (2,3)

$f\u2018\left(a\right)=\underset{h\to 0}{lim}\frac{f(a+h)-f\left(a\right)}{h}$

$=\underset{h\to 0}{lim}\frac{{(a+h)}^{3}-3(a+h)+1-({a}^{3}-3a+1)}{h}$

$=\underset{h\to 0}{lim}\frac{{a}^{3}+3h{a}^{2}+3{h}^{2}a+{h}^{3}-3a-3h+1-{a}^{3}+3a-1}{h}$

$=\underset{h\to 0}{lim}\frac{3h{a}^{2}+3{h}^{2}a+{h}^{3}-3h}{h}$

$=\underset{h\to 0}{lim}(3{a}^{2}+3ha+{h}^{2}-3)$

$=3{a}^{2}+0+0-3$

$=3{a}^{2}-3$

Evaluate the derivative at$a=2$

$f\u2018\left(2\right)=3{\left(2\right)}^{2}-3=9$

$f\u2018\left(x\right)$ is the slope of the tangent line to $f\left(x\right)$ at x, so we have $m=9$ at point (2, 3). Plug into the point-slope line equation

$y-{y}_{1}=m(x-{x}_{1})$

$y=m(x-{x}_{1})+{y}_{1}$

$=9(x-2)+3$

$=9x-18+3$

$=9x-15$

Result

$=9x-15$

Evaluate the derivative at

Result

Maked1954

Beginner2021-11-21Added 17 answers

The given equation of the curve is

We have to find the tangent line to the curve

The point slope form of the tangent line at point

Where

So, slope of the tangent line is given by:

Thus, slope

Now, substitute

Hence, required equation of the tangent line to the curve