Author Topic: Turning point (Read 7572 times) Tweet Share

Hielly · « **Reply #16 on:** January 01, 2010, 04:42:03 pm »

Quote from: Ilovemathsmeth on January 01, 2010, 02:05:40 pm

Well for something like a parabola, you could change to turning point form and THEN work out the coordinates of the turning point. You could also use the x = (-b)/2a rule too.

However for something like a cubic, you've got to use calculus. Or calculator

Hey how would you find the TP of a cubic using calculus? im trying to figure the TP of (x^2)/4*(5-x).

given that two points on the line is (2,3) and (5,0).

Thanks

brightsky · « **Reply #17 on:** January 01, 2010, 04:55:08 pm »

Let $\frac {dy}{dx} = 0$ . The turning points of any graph is the graph's stationary point, that is, when the gradient = 0.

brightsky · « **Reply #18 on:** January 01, 2010, 04:58:24 pm »

Which is the correct interpretation of your function:

$\frac {x^2}{4(5-x)}$

Or....

$\frac {x^2}{4} (5 -x) = \frac {x^2(5-x)}{4}$

Hielly · « **Reply #19 on:** January 01, 2010, 04:59:38 pm »

This:
$\frac {x^2}{4} (5 -x) = \frac {x^2(5-x)}{4}$

brightsky · « **Reply #20 on:** January 01, 2010, 05:20:18 pm »

Ahh ok.

Then:

$f(x) = \frac {5x^2 -x^3}{4}$

$\frac {dy}{dx} = \frac {d}{dx} (\frac {1}{4} (5x^2 - x^3))$

Factor out constants and derive terms one by one:

$\frac {dy}{dx} = \frac {1}{4} (\frac {d}{dx}(5x^2 - x^3))$

$\frac {dy}{dx} = \frac {1}{4} (5(\frac {d}{dx}(x^2) - \frac {d}{dx}(x^3))$

$\frac {dy}{dx} = \frac {1}{4} (5(2x) - (3x^2))$

$\frac {dy}{dx} = \frac {1}{4} (10x - 3x^2)$

For gradient = 0, dy/dx = 0, so:

$\frac {1}{4} (10x - 3x^2) = 0$

$10x - 3x^2 = 0$

$-x (3x - 10) = 0$

So $x = 0$ or $x = \frac {10}{3}$

Substitute the x-values back into the equation:

For x = 0, y = 0.

For x = 10/3,

$y = \frac {(\frac {10}{3})^2 (5 - \frac {10}{3})} {4} = \frac {125}{27}$

So the turning points are at (0,0) and (10/3, 125/27).

Ilovemathsmeth · « **Reply #21 on:** January 01, 2010, 05:49:19 pm »

I suck at using LaTex

Yep so find dy/dx. In above case, probably expand bracket then differentiate.

Find dy/dx = 0. Those are the x-values of the points of zero gradient. MAKE SURE you do a gradient chart - for some cubics, it is actually a stationary point of inflexion which is NOT a turning point. Then substitute into the original function to find y-coordinates.

brightsky · « **Reply #22 on:** January 01, 2010, 05:51:05 pm »

Quote from: Ilovemathsmeth on January 01, 2010, 05:49:19 pm

...for some cubics, it is actually a stationary point of inflexion which is NOT a turning point.

Woops. Forgots about that! :p

brightsky · « **Reply #23 on:** January 01, 2010, 05:54:37 pm »

btw, what's a gradient chart? also, how do you draw them?

Hielly · « **Reply #24 on:** January 01, 2010, 08:53:05 pm »

Okay thanks, how about the TP of $x^4 -4x^2 +4$ ? What's the easiest way? , besides the use of calc?

brightsky · « **Reply #25 on:** January 01, 2010, 09:08:43 pm »

It's basically the same.
1. Find the derivative of the function through differentiation.

$f(x) = x^4 - 4x^2 + 4$

$\frac {dy}{dx} = 4x^3 - 8x$

2. At the turning point, the gradient is 0, so let dy/dx = 0.

$\frac {dy}{dx} = 4x^3 - 8x = 0 \implies x^3 - 2x = 0 \implies x (x^2 - 2) = 0$

By the null factor law, that means:

$x = 0$ or $x^2 - 2 = 0$

When $x^2 - 2 = 0$ , by the quadratic equation:

$x = \frac {-(0) \pm \sqrt {(0)^2 - 4(1)(-2)}}{2(1)}$

$x = \frac {\pm\sqrt {8}}{2} = \frac {\pm2\sqrt{2}}{2} = \pm\sqrt {2}$

Hence, $x = 0$ or $x = \sqrt {2}$ or $x = -\sqrt {2}$

3. Substitute the x-values into the original equation.

When $x = 0$ , $y = 4$ .

When $x = \sqrt {2}$ , $y = 0$ .

When $x = -\sqrt {2}$ , $y = 0$ .

I think that's right..

Edit: Because your graph is a quartic, hence all the solutions stated are turning points.

brightsky · « **Reply #26 on:** January 01, 2010, 09:13:44 pm »

Quote from: Hielly on January 01, 2010, 08:53:05 pm

Okay thanks, how about the TP of $x^4 -4x^2 +4$ ? What's the easiest way? , besides the use of calc?

I think that the best/simpliest/most straight-forward way to find turning points in any function is through the use of the method above.

kamil9876 · « **Reply #27 on:** January 01, 2010, 09:41:32 pm »

another way:

$x^4-4x^2+4=(x^2-2)^2=(x-\sqrt{2})^2(x+\sqrt{2})^2$

Which readily tells you turning points.

As a challenge: Prove WITHOUT using calculus/differentiation that rule about turning points: that the turning point happens at $x=a$ if we can find a term $(x-a)^2$ in our polynomial as a factor (without any other $(x-a)$ factors present)

brightsky · « **Reply #28 on:** January 01, 2010, 09:44:29 pm »

Lolol, listen to kamil. Factorisation, if it's readily seen, can be wayyyyy quicker than going through millions of steps of calculus and whatnot.

Ilovemathsmeth · « **Reply #29 on:** January 01, 2010, 11:27:16 pm »

A gradient chart shows the sign of the slope either side of the point of zero gradient.

Download Word Doc. to see how it should be set out.

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