Answer:
[tex]\displaystyle C=\left(\frac{26}{9}, 2\right)[/tex]
Step-by-step explanation:
We need to find the equation of the tangent lines of Points A and B.
Differentiate the equation:
[tex]\displaystyle \frac{dy}{dx}=3x^2-3[/tex]
Point A has an x-coordinate of -1. Hence, the slope of its tangent line is:
[tex]\displaystyle \frac{dy}{dx}\Big|_{x=-1}=3(-1)^2-3=0[/tex]
Find the y-coordinate of Point A using the original equation:
[tex]y(-1)=(-1)^3-3(-1)=2[/tex]
Hence, Point A is (-1, 2).
Thus, the tangent line at Point A is:
[tex]y-2=0(x-(-1))[/tex]
Simplify:
[tex]y=2[/tex]
Point B has an x-coordinate of 4. Hence, the slope of its tangent line is:
[tex]\displaystyle \frac{dy}{dx}\Big|_{x=4}=3(4)^2-3=45[/tex]
Find the y-coordinate of Point B:
[tex]y(4)=(4)^3-3(4)=52[/tex]
Thus, Point B is at (4, 52).
So, the tangent line at Point B is:
[tex]y-52=45(x-4)[/tex]
Simplify:
[tex]y=45x-128[/tex]
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:
[tex]2=45x-128[/tex]
We acquire:
[tex]\displaystyle x=\frac{26}{9}[/tex]
Since one of our equations is y = 2, the y-coordinate is 2.
Hence, Point C is:
[tex]\displaystyle C=\left(\frac{26}{9}, 2\right)[/tex]
Answer:
Step-by-step explanation:
y=x³-3x
[tex]\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[/tex]
so coordinates of C are (3,2)
