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

Guest Jan 5, 2015

#3**+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.

Melody 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

Guest Jan 5, 2015

#3**+5 **

Best Answer

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.

Melody Jan 5, 2015