Graphing the derivative HSC question (2 Viewers)

BlueGas · Oct 1, 2015

Okay so this is the first derivative, now what do I do?

InteGrand · Oct 1, 2015

BlueGas said:
Okay so this is the first derivative, now what do I do?

$Now sketch the derivative of that. Call that graph you drew $y=g(x)$ (clearly, $g$ is just $f^\prime$, i.e. the derivative of the original given function). We want to sketch the graph of $y=g^\prime (x)$ (which is $f''(x)$). Just apply what you learnt about how to sketch the first derivative of a graph to this new graph of $y=g(x)$.$

kawaiipotato · Oct 1, 2015

Label that new curve as g (x) to avoidconfusion and then repeat the steps you did before to find g'(x) (which is the second derivative of the original curve)

BlueGas · Oct 1, 2015

Same steps? Okay lemme try this.

- At x = -2 there's a minimum, so g'(-2)=0? Because whatever stationary points there are they become intercepts
- To the left of x = -2 g''(x) > 0 so it'll start from the bottom of the graph
- At x = 1 there's a maximum so g'(1)=0 (same thing from the first point)
- At x = 3 I don't really know what to do for that I think it's not counted because it's already an "intercept"
- So basically this will look like a parabola? An upside down one?

InteGrand · Oct 1, 2015

BlueGas said:
will look like a parabola? An upside down one?

$Yes, like I said before, the graph should look like a quadratic, and it would be concave down because as $x\to \pm \infty$, $g^\prime (x) \to -\infty$, which would not work if the parabola were upward facing, as that'd imply that $g^\prime (x) \to +\infty$ instead.$

BlueGas · Oct 1, 2015

InteGrand said:
$Yes, like I said before, the graph should look like a quadratic, and it would be concave down because as $x\to \pm \infty$, $g^\prime (x) \to -\infty$, which would not work if the parabola were upward facing, as that'd imply that $g^\prime (x) \to +\infty$ instead.$

So was my steps/working out correct?

InteGrand · Oct 1, 2015

BlueGas said:
Same steps? Okay lemme try this.

- At x = -2 there's a minimum, so g'(-2)=0? Because whatever stationary points there are they become intercepts
- To the left of x = -2 g''(x) > 0 so it'll start from the bottom of the graph
- At x = 1 there's a maximum so g'(1)=0 (same thing from the first point)
- At x = 3 I don't really know what to do for that I think it's not counted because it's already an "intercept"
- So basically this will look like a parabola? An upside down one?

BlueGas said:
So was my steps/working out correct?

$The point $x=3$ is of no significance for the graph of $y=g^\prime (x)$, as the derivative of $g$ there is nothing special. One point of significance you missed is the $x$ value of the midpoint of the two local extrema of the graph of $y=g(x)$ (call this $x$ value $x^*$). At $x^*$, we will actually have an inflexion point (assuming the graph is a cubic), and this is where the vertex of the quadratic $y=g^\prime (x)$ would be. One fact about cubic functions is that any cubic function has exactly one inflexion point (if the cubic has two turning points, then this inflexion point's $x$-value is midway between those of the turning points), and the graph of any cubic function is symmetric about its inflexion point (``symmetric'' in a similar way to how the graph of an odd function is symmetric about the origin).$

BlueGas · Oct 1, 2015

And I think the point of inflexion is at one, but at the same time there's a maximum too, so I wonder what happens then.

InteGrand · Oct 1, 2015

BlueGas said:
And I think the point of inflexion is at one, but at the same time there's a maximum too, so I wonder what happens then.

$The point of inflexion is for the function $g$, whereas the maximum is for the function $g^\prime$ (it's not an inflexion point for $g^\prime$).$

BlueGas · Oct 1, 2015

InteGrand said:
$The point of inflexion is for the function $g$, whereas the maximum is for the function $g^\prime$ (it's not an inflexion point for $g^\prime$).$

So when drawing the derivative of g(x), I ignore the point of inflexion and just look at the maximum at x = 1?

InteGrand · Oct 1, 2015

BlueGas said:
So when drawing the derivative of g(x), I ignore the point of inflexion and just look at the maximum at x = 1?

$That point of inflexion \emph{of g} is the maximum of $g^\prime$. And $g^\prime$ has no inflexion points if it is a quadratic (no quadratics do).$

BlueGas · Oct 1, 2015

InteGrand said:
$That point of inflexion \emph{of g} is the maximum of $g^\prime$. And $g^\prime$ has no inflexion points if it is a quadratic (no quadratics do).$

OHHHHHHHHHHHHH okay now I get it lol, thanks alot man for your help, I really appreciate it!

Graphing the derivative HSC question (2 Viewers)

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