Question

The tangents to the curve with equation y = x^3-3x at the points A and B with x coordinates −1 and 4 respectively meet at the point C. Find the coordinates of the point C.

Answer 1

Answer:

$\displaystyle C=\left(\frac{26}{9}, 2\right)$

Step-by-step explanation:

We need to find the equation of the tangent lines of Points A and B.

Differentiate the equation:

$\displaystyle \frac{dy}{dx}=3x^2-3$

Point A has an x-coordinate of -1. Hence, the slope of its tangent line is:

$\displaystyle \frac{dy}{dx}\Big|_{x=-1}=3(-1)^2-3=0$

Find the y-coordinate of Point A using the original equation:

$y(-1)=(-1)^3-3(-1)=2$

Hence, Point A is (-1, 2).

Thus, the tangent line at Point A is:

$y-2=0(x-(-1))$

Simplify:

$y=2$

Point B has an x-coordinate of 4. Hence, the slope of its tangent line is:

$\displaystyle \frac{dy}{dx}\Big|_{x=4}=3(4)^2-3=45$

Find the y-coordinate of Point B:

$y(4)=(4)^3-3(4)=52$

Thus, Point B is at (4, 52).

So, the tangent line at Point B is:

$y-52=45(x-4)$

Simplify:

$y=45x-128$

Point C occurs at the intersections of the tangent lines of Points A and B. Set the two equations equal to each other and solve for x:

$2=45x-128$

We acquire:

$\displaystyle x=\frac{26}{9}$

Since one of our equations is y = 2, the y-coordinate is 2.

Hence, Point C is:

$\displaystyle C=\left(\frac{26}{9}, 2\right)$

Answer 2

Answer:

Step-by-step explanation:

y=x³-3x

$\frac{dy}{dx} =3x^2-3\\when x=-1,\frac{dy}{dx} =3(-1)^2-3=3-3=0\\when x=-1\\y=(-1)^3-3(-1)=-1+3=2\\eq. ~of~line~through ~(-1,2)~with~slope=0~is\\y-2=0(x+1)\\or y=2\\when~x=4,\\\frac{dy}{dx} =3(4)^2-3=48-3=45\\when x=4\\y=4^3-3(4)=64-12=52\\so~eq.~ of~line~is~\\y-52 =45(x-4)\\y-52=45x-180\\or ~y=45x-180+52\\or\\y=45x-128\\when ~y=2\\2=45 x-128\\45 x=2+128\\45 x=130\\x=130/45=3$

so coordinates of C are (3,2)