Find the slope of the normal line to the function f(x) = sin(cos 5x) at the point x=п/10
+118609 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.
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
+118609 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.
