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

Question

Find an equation of the tangent line to the curve at the given point. y=x3−3x+1, (2,3)

Antum1978 · Accepted Answer

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

f (x) = x^{3} - 3 x + 1

, (2,3)

f ‘ (a) = lim_{h \to 0} \frac{f (a + h) - f (a)}{h}

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

= lim_{h \to 0} \frac{a^{3} + 3 h a^{2} + 3 h^{2} a + h^{3} - 3 a - 3 h + 1 - a^{3} + 3 a - 1}{h}

= lim_{h \to 0} \frac{3 h a^{2} + 3 h^{2} a + h^{3} - 3 h}{h}

= lim_{h \to 0} (3 a^{2} + 3 h a + h^{2} - 3)

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

= 3 a^{2} - 3

Evaluate the derivative at

a = 2

f ‘ (2) = 3 {(2)}^{2} - 3 = 9

f ‘ (x)

is the slope of the tangent line to

f (x)

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

= 9 x - 18 + 3

= 9 x - 15

Result

= 9 x - 15

Maked1954 · Accepted Answer

The given equation of the curve is y=x3−3x+1. We have to find the tangent line to the curve y=x3−3x+1 at point (2, 3). The point slope form of the tangent line at point (x1,y1) is given by: (y−y1)=m(x−x1) Where m=dydx∣(x1−y1) the slope of the tangent line at point (x1,y1) which is equal to derivative ofthe curve y at point (x1,y1). So, slope of the tangent line is given by: m=dydx∣(2,3) =d(x3−3x+1)dx∣(2,3) =(3x2−3)∣(2,3) =3×22−3 {On replacing x with 2} =12−3 =9 Thus, slope m=9 Now, substitute m=9 and given point (x1,y1)=(2,3) in the general form of the line (y−y1)=m(x−x1). =(y−3)=9(x−2) y−3=9x−18 9x−y−15=0 Hence, required equation of the tangent line to the curve y=x3−3x+1 at point (2, 3) is 9x−y−15=0.

Find an equation of the tangent line to the curve

Answered question

Answer & Explanation

New Questions in Other