+0

# Find the slope of the normal line to the function f(x) = sin(cos 5x) at the point x=п/10

0
293
3

Find the slope of the normal line to the function f(x) = sin(cos 5x) at the point x=п/10

Jan 5, 2015

#3
+102763
+5

The tangent line to the curve y = x^3 at the point
A(2, 8) intersects the curve at another point

$$\\y=x^3\\ y'=3x^2\\ when x=2\\ y'=3*4=12\\  so the gradient of the tangent is 12\\ equation of the tangent is\\ y=12x+b\\ sub in (2,8)\\ 8=24+b\\ b=-16\\ y=12x-16\\ \mbox{So you need to find the intersection of }\\ y=12x-16 \quad and \quad y=x^3\\\\ x^3=12x-16\\ x^3-12x+16=0\\ I know that (2,8) is one solution so one of the factors is (x-2)\\ I now did an algebraic division and found the other factor to be x^2+2x-8\\$$

$$\\So\\ x^3-12x+16=0\\ (x-2)(x^2+2x-8)=0\\ (x-2)(x-2)(x+4)=0\\ So the intersections occur at x=2 and at x=-4\\ when\;\; x=-4 \qquad y=(-4)^3=-64\\ so\\ the other intersection will be at (-4,-64)\\$$

and just to make sure it was right I draw the graph.

https://www.desmos.com/calculator/hsbvq84aub

Jan 5, 2015

#1
+5

thanks guys! I got it!

Jan 5, 2015
#2
0

The tangent line to the curve y = x3 at the point
A(2, 8) intersects the curve at another point
B(x0, y0). Find x0.  And this is another question for you guys... sincerely

Jan 5, 2015
#3
+102763
+5

The tangent line to the curve y = x^3 at the point
A(2, 8) intersects the curve at another point

$$\\y=x^3\\ y'=3x^2\\ when x=2\\ y'=3*4=12\\  so the gradient of the tangent is 12\\ equation of the tangent is\\ y=12x+b\\ sub in (2,8)\\ 8=24+b\\ b=-16\\ y=12x-16\\ \mbox{So you need to find the intersection of }\\ y=12x-16 \quad and \quad y=x^3\\\\ x^3=12x-16\\ x^3-12x+16=0\\ I know that (2,8) is one solution so one of the factors is (x-2)\\ I now did an algebraic division and found the other factor to be x^2+2x-8\\$$

$$\\So\\ x^3-12x+16=0\\ (x-2)(x^2+2x-8)=0\\ (x-2)(x-2)(x+4)=0\\ So the intersections occur at x=2 and at x=-4\\ when\;\; x=-4 \qquad y=(-4)^3=-64\\ so\\ the other intersection will be at (-4,-64)\\$$

and just to make sure it was right I draw the graph.

https://www.desmos.com/calculator/hsbvq84aub

Melody Jan 5, 2015