Graphing the derivative HSC question (1 Viewer)

BlueGas · Sep 10, 2015

I need help with this question, I have no clue how to start so I need to know the basics.

Drsoccerball · Sep 10, 2015

BlueGas said:
I need help with this question, I have no clue how to start so I need to know the basics.

When differentiating a function turning points become x intercepts and inflection points become turning point. So considering this in the domain of
$\leq -2$ the function start from positive infinity y axis (Since gradient appears to be infinitely positive) and is going down till -2. So as mentioned before stationary points become intercepts and inflection points become turning points therefore there would be a minimum stationary point at x=-2. So the graph then increases untill a min turning point at x=1. At x=3 it crosses the x axis and continues downwards

BlueGas · Oct 1, 2015

Drsoccerball said:
When differentiating a function turning points become x intercepts and inflection points become turning point. So considering this in the domain of
$\leq -2$ the function start from positive infinity y axis (Since gradient appears to be infinitely positive) and is going down till -2. So as mentioned before stationary points become intercepts and inflection points become turning points therefore there would be a minimum stationary point at x=-2. So the graph then increases untill a min turning point at x=1. At x=3 it crosses the x axis and continues downwards

Just a quick bump, since inflection points become stationary/turning points, how'd you know that x= -2 it's a minimum and at x= 1 there's another minimum?

Drsoccerball · Oct 1, 2015

BlueGas said:
Just a quick bump, since inflection points become stationary/turning points, how'd you know that x= -2 it's a minimum and at x= 1 there's another minimum?

Because you look at the gradient before and after it:
Its increasing and the after the point it decrease. The only point that does that is a max turning point

InteGrand · Oct 1, 2015

BlueGas said:
Just a quick bump, since inflection points become stationary/turning points, how'd you know that x= -2 it's a minimum and at x= 1 there's another minimum?

$At $x=-2$, $f''(x)=0$, and just to the left of $x=-2$, we have $f''(x)<0$ (easy to see from the concavity of the graph), and just to the right of $x=-2$, we have $f''(x)>0$ (again see the concavity), so at $x=-2$, $f^\prime$ must have a local \emph{minimum} (this is the \textsl{first derivative test} applied to the function $f^\prime$, whose first derivative is $f''$). Also, this minimum is $0$, since $f^\prime (-2)=0$.$

$The same logic applies to the point $x=1$. The only difference is that the local minimum of $f^\prime$ here is clearly a positive number, as $f^\prime (1)>0$ (as the slope there is positive).$

BlueGas · Oct 1, 2015

Drsoccerball said:
Because you look at the gradient before and after it:
Its increasing and the after the point it decrease. The only point that does that is a max turning point

Wait so it's a maxiumum? You said max turning point? I thought it was a minimum?

BlueGas · Oct 1, 2015

Drsoccerball said:
When differentiating a function turning points become x intercepts and inflection points become turning point. So considering this in the domain of
$\leq -2$ the function start from positive infinity y axis (Since gradient appears to be infinitely positive) and is going down till -2. So as mentioned before stationary points become intercepts and inflection points become turning points therefore there would be a minimum stationary point at x=-2. So the graph then increases untill a min turning point at x=1. At x=3 it crosses the x axis and continues downwards

Also you said at x = 1 there's a minimum, but the actual answer has x = 1 as a maximum.

BlueGas · Oct 1, 2015

InteGrand said:
$At $x=-2$, $f''(x)=0$, and just to the left of $x=-2$, we have $f''(x)<0$ (easy to see from the concavity of the graph), and just to the right of $x=-2$, we have $f''(x)>0$ (again see the concavity), so at $x=-2$, $f^\prime$ must have a local \emph{minimum} (this is the \textsl{first derivative test} applied to the function $f^\prime$, whose first derivative is $f''$). Also, this minimum is $0$, since $f^\prime (-2)=0$.$

$The same logic applies to the point $x=1$. The only difference is that the local minimum of $f^\prime$ here is clearly a positive number, as $f^\prime (1)>0$ (as the slope there is positive).$

Okay so to the left of x = -2 is a max and to the right is a min, so how did you conclude that it's a minimum when sketching the derivative?

InteGrand · Oct 1, 2015

BlueGas said:
Okay so to the left of x = -2 is a max and to the right is a min, so how did you conclude that it's a minimum when sketching the derivative?

Sorry, yeah, local maximum for f' at x = 1, because the derivative of f' is positive to the left of it (that is, f"(x) > 0 just to the left of 1, as the curve is concave up there) and the derivative of f' is negative to the right of x = 1 (as the concanvity there is concave down).

InteGrand · Oct 1, 2015

BlueGas said:
Okay so to the left of x = -2 is a max and to the right is a min, so how did you conclude that it's a minimum when sketching the derivative?

$If a smooth function $g$ has $g^\prime (a) = 0$, and $g^\prime (x) < 0$ just to the left of $a$ and $g^\prime (a)> 0$ just to the right of $a$, then at $a$, $g$ has a local minimum (flip all the inequality signs and instead it is a local maximum). Apply this to the function $f^\prime$, that is, take $g = f^\prime$, so $g^\prime = f''$. I am pretty sure you would have learnt this test (first derivative test) when learning about optimisation of functions.$

BlueGas · Oct 1, 2015

InteGrand said:
$If a smooth function $g$ has $g^\prime (a) = 0$, and $g^\prime (x) < 0$ just to the left of $a$ and $g^\prime (a)> 0$ just to the right of $a$, then at $a$, $g$ has a local minimum (flip all the inequality signs and instead it is a local maximum). Apply this to the function $f^\prime$, that is, take $g = f^\prime$, so $g^\prime = f''$. I am pretty sure you would have learnt this test (first derivative test) when learning about optimisation of functions.$

That's the problem, my teacher NEVER touched on this topic, it's unbelievable, as a result I'm having a hard time understanding this graphing the derivative and so on. I only have a vague idea on this.

So far what I've learnt, not only from this question, but other questions is that:
f(x) is if the point is positive or negative, basically if the graph is above the x axis or not, for example at say (-2, 3) f(x) > 0 however if it was (-2, -2) it would instead be f(x) < 0
f'(x) is the slope/gradient, if it's positive or negative
f''(x) is if it's a max or a min

BlueGas · Oct 1, 2015

Also how do I know where the graph starts from? From top or bottom?

kawaiipotato · Oct 1, 2015

BlueGas said:
Also how do I know where the graph starts from? From top or bottom?

On the most left, f(x) appears to be increasing, that is, f'(x) >0. The concavity of f (x) is also negative, so f''(x) <0. That means for the f'(x) curve, the derivative (ie. The gradient) of that curve is also negative since f"(x)<0. So we can say that the f '(x) starts positive (above the x axis) and decreases down

BlueGas · Oct 1, 2015

kawaiipotato said:
On the most left, f(x) appears to be increasing, that is, f'(x) >0. The concavity of f (x) is also negative, so f''(x) <0. That means for the f'(x) curve, the derivative (ie. The gradient) of that curve is also negative since f"(x)<0. So we can say that the f '(x) starts positive (above the x axis) and decreases down

Ah okay, so when f''(x) < 0, then the gradient is a negative and starts from the top?

kawaiipotato · Oct 1, 2015

BlueGas said:
Ah okay, so when f''(x) < 0, then the gradient is a negative and starts from the top?

If f"(x)<0 then the gradient of y = f'(x) is negative. For the y = f (x), it only tells us that it has a negative concavity (think, a concave down parabola). It only starts at the top for the y=f'(x) curve because for y=f (x), the gradient (and hence f'(x)) is positive.
.
.
.
So it starts at top b/c gradient of f (x) is positive and it has a negative gradient because f''(x) is negative

BlueGas · Oct 1, 2015

I'm getting a much better idea on this now, thanks alot guys! I can't wait to stop posting questions for help after the HSC, it's too tiring

BlueGas · Oct 1, 2015

Since this is only finding the first derivative, what are the steps I should take to find the second derivative?

InteGrand · Oct 1, 2015

BlueGas said:
Since this is only finding the first derivative, what are the steps I should take to find the second derivative?

You can try graphing the first derivative, and then use this graph to graph the second derivative.

kawaiipotato · Oct 1, 2015

What InteGrand said.

But if you're able to, you might be able to draw the second derivative from the y = f(x) graph by assessing the concavities at each point.
eg. from x = -infinity to x = -2,
the concavity is highly negative, until it reaches x = -2 where there's an inflexion point, showing us that f''(x) = 0. Then after this point, the concavity is positive.
So you can represent this on a graph, starting at x=-infinity, f''(x) = very negative and it goes up until x= -2 where it crosses the x-axis and continues going up and becomes positive because the concavity of f(x) after x = -2 is positive

InteGrand · Oct 1, 2015

Also note that the given graph looks like that of a quartic (fourth degree polynomial), so we would expect the first derivative graph to look like that of a cubic, and the second derivative graph to look like that of a quadratic.

Graphing the derivative HSC question (1 Viewer)

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