# Thread: [SOLVED] Curve Sketching

1. ## [SOLVED] Curve Sketching

a)Find the equation of the tangent at the point (t, $t^3$) on the curve $y=x^3$.Find also the coordinates of the point where this tangent meets the curve again.

b) For the curve, $y=(4/x^2)+x$ , show the following ,
i) the entire curve lies above the line y=x

Red colour fonts are parts which I need help, thanks in advance!

2. a) $y' = 3x^2$, so the slope at point x=t is $3t^2$. Now we have a point and a slope, thus we're able to make a line.

$y-t^3=3t^2(x-t)$, plugging in the info into the point-slope equation for a line. Simplify that out and you'll get your tangent equation.

To see where these two lines meet again, set them equal to each other and solve.

3. Originally Posted by Jameson
a) $y' = 3x^2$, so the slope at point x=t is $3t^2$. Now we have a point and a slope, thus we're able to make a line.

$y-t^3=3t^2(x-t)$, plugging in the info into the point-slope equation for a line. Simplify that out and you'll get your tangent equation.

To see where these two lines meet again, set them equal to each other and solve.
Set these two equations together?
$y=x^3$ and $y=3t^2(x-t)+t^3$

4. Yessir.

5. $y=(4/x^2)+x$

b) Assume that the curve $y=(4/x^2)+x$ and the line y=x intersects. Therefore:
$x = (4/x^2)+x$

Canceling x, you would get the ridiculous statement that:
$0 = \frac{4}{x^2}$

$0 = 4$

There is a contradiction, therefore our original assumption is not true.

EDIT: Continue with the next post below.

6. Originally Posted by Jameson
Yessir.
I'm sorry to trouble you again, but I cannot solve the equation for x.

7. Originally Posted by Chop Suey
$y=(4/x^2)+x$

b) Assume that the curve $y=(4/x^2)+x$ and the line y=x intersects. Therefore:
$x = (4/x^2)+x$

Canceling x, you would get the ridiculous statement that:
$0 = \frac{4}{x^2}$

$0 = 4$

There is a contradiction, therefore our original assumption is not true.

EDIT: I didn't read your question correctly. I will edit again.

Isn't that an inclined asymptote for the curve?
I can't explain why it is above the line y=x.

8. No problem

So $x^3=3t^2(x)-3t^3+t^3$

Simply $x^3=3t^2(x)-2t^3$

Group and factor $x(x^2-3t^2)=-2t^3$

Made a mistake... working on it

9. Originally Posted by ose90
Isn't that an inclined asymptote for the curve?
I can't explain why it is above the line y=x.
$f(x) = \frac{4}{x^2} + x$

$g(x) = x$

Let's assume that f(x) < g(x). So that:
$\frac{4}{x^2} + x < x$

But if you cancel the x, you see that:
$\frac{4}{x^2}<0$

$4<0$

Which is a contradiction.

$\therefore f(x) > g(x)$

The reason why I didn't put the equality sign is because I showed that they do not intersect.

10. Originally Posted by Chop Suey
$f(x) = \frac{4}{x^2} + x$

$g(x) = x$

Let's assume that f(x) < g(x). So that:
$\frac{4}{x^2} + x < x$

But if you cancel the x, you see that:
$\frac{4}{x^2}<0$

$4<0$

Which is a contradiction.

$\therefore f(x) > g(x)$

The reason why I didn't put the equality sign is because I showed that they do not intersect.
Thanks a lot

11. Originally Posted by Jameson
No problem

So $x^3=3t^2(x)-3t^3+t^3$

Simply $x^3=3t^2(x)-2t^3$

Group and factor $x(x^2-3t^2)=-2t^3$

Made a mistake... working on it
Thanks for helping me. I have tried to solve it for many times but still can't get the value of x. Maybe there is no solution.

12. You're welcome. I think there's a trick I'm not seeing. Cubics that have no x^2 term have ways of being solved. Either someone else will see it or I'll solve it soon.

13. Notice that x-t is a factor, and by polynomial division, you can reduce it to:
$(x-t)(x^2+tx-2t^2)=0$

And it can be factored even further to:
$(x-t)(x-t)(x+2t)=0$

14. thanks for your help!