Calculus Help

- anonymous

Calculus Help

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- anonymous

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- DanJS

they give you the first derivative of g. g ' (x)
maybe start off by figuring out the function g(x) and thsecond derivative g ''(x)

- Rizags

The key critical values of a function g(x) occur where g'(x)=0 (Stationary Points), or g'(x) is undefined.

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## More answers

- anonymous

I'm not sure how to "un-derive" but the second derivative I got was f''(x)=\[x^2-4x+16\]

- DanJS

well just g '' don't think you need g

- anonymous

oops i forgot the denominator. x^2-4x+16/(x-2)^2

- anonymous

for number 1, I set the first derivative to 0 and DNE. and I got -4,2,4

- Rizags

those should be your critical values. Concave up/down is determined by the sign of g''(x) at the key point. Simply evaluate g''(-4), g''(2), g''(4). If g''(x)>0, the graph is concave up there. If g''(x)<0, the graph is concave down there.

- Rizags

nevermind, sorry didn't read the question, you don't need that

- anonymous

I thought I had to set the second derivative to 0 and DNE to find the inflection points. And those determine concave up or down? Thats number 3

- DanJS

|dw:1449622493694:dw|

- Rizags

g''(x)=0 and g''(x) DNE gives inflection points, yes. If g''(x)<0 for values in the interval between inflection points, then the graph of g is concave DOWN for that interval. Similarly, id g''(x)>0 for values in the interval between inflection points, the the graph of g is concave UP on that interval.

- DanJS

pick a value inside each interval, test it in g '(x), to see if you get + or -, inc or dec

- anonymous

I had a problem setting the second derivative = to 0 and DNE. I couldn't get any points

- anonymous

@DanJS I got decreasing, increasing, decreasing, decreasing

- DanJS

ok, the points are the critical points for #1
#2, say that for each interval (-inf,-4), (-4,-2)...

- DanJS

wait, x=5 gives + number, that is increasing

- anonymous

I made the chart
|dw:1449622831451:dw|
youre right about the last one

- DanJS

this is all #2.
if g(x) changes from iincreasing to decreasing --that is a local Max
decreasing to incrasing --- local Min

- DanJS

yeah sorry , you said
I got decreasing, increasing, decreasing, decreasing

- anonymous

so the local max is at 2?
and the local min at -4 and 4?

- DanJS

for #2 also, may want to say something about the derivative being the slope of a tangent to the function at a given point... the slopes of those tangents change from + to 0 to -, at each of those critical values

- DanJS

yep

- anonymous

alright, do you know how to do 3? I got stuck on this one

- DanJS

yeah, concavity has to do with second derivative , g ''
similar process, first find when that is 0 and set up the intervals

- anonymous

2 would be a possible inflection point because it DNE. I set the equation = to 0 and I didn't get any values from it

- anonymous

I got the g''(x)= \[(x^2-4x+16)/(x-2)^2\]
I'm left with a quadratic equation that results in imaginary numbers

- anonymous

wait, isn't it concave down when the slope of the tangent line is decreasing? and concave up when its increasing?

- DanJS

sorry, here

- DanJS

ok, see what the critical points do from the first derivative

- anonymous

I got concave down at x<-4 and 2

- anonymous

I'm not sure if the concavity can be determined from the first derivative. I think I have to find the possible inflection points, but that's what I'm having trouble finding

- DanJS

evaluate g ''(x) for each of the critical values, x=-4, 2 , 4

- anonymous

plug in the critical points, or using numbers other than them, like in the chart I made?

- anonymous

for x=-4, I got 4/3

- anonymous

x=2 DNE and x=4 I got 4

- DanJS

x=2 is said to be undefined for the function, there is something there, a hole or cusp or whatever
we had that as a crit point already and tested the inc/dec there, it changes
need to also look at the concavity also

- anonymous

So I'm confused on how to find concavity. I didn't know you could use the critical points. I thought you had to find the possible inflection points then evaluate in terms of the second derivative

- anonymous

@Rizags

- anonymous

on number 3

- anonymous

DanJs

- anonymous

Can you help me on my problem? Would be much appreciated.

- anonymous

@DanJS do you understand 4 and 5?

- Rizags

Concavity can be determined using either the first or second derivative, with the second being markedly easier.

- anonymous

Can you explain in context to this problem how to find it using the second derivative. I'm stuck setting the equation to 0

- Rizags

could you post the equation for the second derivative here?

- anonymous

\[(x^2-4x+16)/(x-2)^2\]

- anonymous

I think i did that right. I'll double check

- anonymous

yeah I got the same thing again

- Rizags

Okay, so the issue here is that g''(x) never =0, but g''(x) CAN be DNE, when x=2, nevertheless, I think we're going to have to use the first derivative test

- Rizags

Critical points are -4, 2, and 4 as we said

- anonymous

ok I have plugged them in and got 4/3, DNE, 4

- Rizags

right, here's the weird thing about this graph. although it does change concavity, it doesn't actually have a point of inflection

- DanJS

at that x=2, the function does not exist, but the , and changes increasing to decreasing

- DanJS

it is a hole in the graph, or a cusp, piecewise defined, i think

- anonymous

so its concave down at -4 and 4? because f''(c)>0?

- Rizags

I'm actually making a really dumb mistake here. If you plug in g''(-4) and g''(4), the answers will both be positive, signaling that the entire graph of g is concave up, however the issue in point is that g is not differentiable at x=2 which is why this oddity is happening.

- DanJS

|dw:1449628123629:dw|

- anonymous

so its not concave down at ant points?

- Rizags

nope, its concave up on (-inf, 2) and (2, inf)

- Rizags

we can't know what occurs at 2 because g isn't differentiable there; its a cusp

- anonymous

so question 3 is a trick question?

- DanJS

yes, like the absolute value piecwise dfined functnon

- DanJS

hole in the graph rather, because here g(2) is not defined in the function

- anonymous

alright so what about 4 and 5? I have no idea what they are even asking about

- Rizags

Evaluate g'(3) to begin. That will be the slope of the tangent at x=3.

- DanJS

this is just a special case, the continuous functions are straight forward

- anonymous

g'(3) is -7?

- Rizags

Here's what we know about the tangent line: its slope is g'(3), and it runs through the point (3, 4). This is sufficient information to calculate it.

- anonymous

y-4=-7(x-3)?

- DanJS

i guess you can say concavity because of the number of inc/dec shifts, that is some higher degree polynomial, it has curvature

- DanJS

and the derivative exists

- Rizags

yea that looks correct. Also, for any concave UP function at a point, the tangent line to the point will always be BELOW its graph, and vice versa

- anonymous

where did you get the 4 from in the point (3,4)

- Rizags

in the problem, it says g(3)=4

- anonymous

ohhh ok

- anonymous

how do i determine if it lies above or below?

- Rizags

for any concave UP function at a point, the tangent line to the point will always be BELOW its graph. this can be seen intuitively by imagining y=x^2. The tangent to any of the points will always be below. Similarly, the converse is true; for any concave DOWN function at a point, the tangent line to the point will always be ABOVE its graph. Again, envision y=-x^2

- anonymous

so since the entire graph is concave up, the tangent line lies below?

- Rizags

correct. But the reasoning is wrong. The entire graph doesn't have to be concave up, it only has to be concave up for that specific x value (in this case x=3)

- anonymous

so i should say: at x=3, the graph is concave up, therefore the tangent line lies below it

- Rizags

correct.

- anonymous

alright thanks so much guys

- Rizags

ya np

- DanJS

welcome

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