lilyta79jd

2021-11-19

Find an equation of the tangent line to the curve at the given point.
$y={x}^{3}-3x+1$, (2,3)

Antum1978

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)
$f‘\left(a\right)=\underset{h\to 0}{lim}\frac{f\left(a+h\right)-f\left(a\right)}{h}$
$=\underset{h\to 0}{lim}\frac{{\left(a+h\right)}^{3}-3\left(a+h\right)+1-\left({a}^{3}-3a+1\right)}{h}$
$=\underset{h\to 0}{lim}\frac{{a}^{3}+3h{a}^{2}+3{h}^{2}a+{h}^{3}-3a-3h+1-{a}^{3}+3a-1}{h}$
$=\underset{h\to 0}{lim}\frac{3h{a}^{2}+3{h}^{2}a+{h}^{3}-3h}{h}$
$=\underset{h\to 0}{lim}\left(3{a}^{2}+3ha+{h}^{2}-3\right)$
$=3{a}^{2}+0+0-3$
$=3{a}^{2}-3$
Evaluate the derivative at $a=2$
$f‘\left(2\right)=3{\left(2\right)}^{2}-3=9$
$f‘\left(x\right)$ 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
$y-{y}_{1}=m\left(x-{x}_{1}\right)$
$y=m\left(x-{x}_{1}\right)+{y}_{1}$
$=9\left(x-2\right)+3$
$=9x-18+3$
$=9x-15$
Result
$=9x-15$

Maked1954

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

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