anonymous
  • anonymous
Calculus Help
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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DanJS
  • 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
  • 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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anonymous
  • anonymous
I'm not sure how to "un-derive" but the second derivative I got was f''(x)=\[x^2-4x+16\]
DanJS
  • DanJS
well just g '' don't think you need g
anonymous
  • anonymous
oops i forgot the denominator. x^2-4x+16/(x-2)^2
anonymous
  • anonymous
for number 1, I set the first derivative to 0 and DNE. and I got -4,2,4
Rizags
  • 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
  • Rizags
nevermind, sorry didn't read the question, you don't need that
anonymous
  • 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
  • DanJS
|dw:1449622493694:dw|
Rizags
  • 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
  • DanJS
pick a value inside each interval, test it in g '(x), to see if you get + or -, inc or dec
anonymous
  • anonymous
I had a problem setting the second derivative = to 0 and DNE. I couldn't get any points
anonymous
  • anonymous
@DanJS I got decreasing, increasing, decreasing, decreasing
DanJS
  • DanJS
ok, the points are the critical points for #1 #2, say that for each interval (-inf,-4), (-4,-2)...
DanJS
  • DanJS
wait, x=5 gives + number, that is increasing
anonymous
  • anonymous
I made the chart |dw:1449622831451:dw| youre right about the last one
DanJS
  • DanJS
this is all #2. if g(x) changes from iincreasing to decreasing --that is a local Max decreasing to incrasing --- local Min
DanJS
  • DanJS
yeah sorry , you said I got decreasing, increasing, decreasing, decreasing
anonymous
  • anonymous
so the local max is at 2? and the local min at -4 and 4?
DanJS
  • 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
  • DanJS
yep
anonymous
  • anonymous
alright, do you know how to do 3? I got stuck on this one
DanJS
  • DanJS
yeah, concavity has to do with second derivative , g '' similar process, first find when that is 0 and set up the intervals
anonymous
  • 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
  • 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
  • anonymous
wait, isn't it concave down when the slope of the tangent line is decreasing? and concave up when its increasing?
DanJS
  • DanJS
sorry, here
DanJS
  • DanJS
ok, see what the critical points do from the first derivative
anonymous
  • anonymous
I got concave down at x<-4 and 2
anonymous
  • 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
  • DanJS
evaluate g ''(x) for each of the critical values, x=-4, 2 , 4
anonymous
  • anonymous
plug in the critical points, or using numbers other than them, like in the chart I made?
anonymous
  • anonymous
for x=-4, I got 4/3
anonymous
  • anonymous
x=2 DNE and x=4 I got 4
DanJS
  • 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
  • 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
  • anonymous
@Rizags
anonymous
  • anonymous
on number 3
anonymous
  • anonymous
DanJs
anonymous
  • anonymous
Can you help me on my problem? Would be much appreciated.
anonymous
  • anonymous
@DanJS do you understand 4 and 5?
Rizags
  • Rizags
Concavity can be determined using either the first or second derivative, with the second being markedly easier.
anonymous
  • 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
  • Rizags
could you post the equation for the second derivative here?
anonymous
  • anonymous
\[(x^2-4x+16)/(x-2)^2\]
anonymous
  • anonymous
I think i did that right. I'll double check
anonymous
  • anonymous
yeah I got the same thing again
Rizags
  • 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
  • Rizags
Critical points are -4, 2, and 4 as we said
anonymous
  • anonymous
ok I have plugged them in and got 4/3, DNE, 4
Rizags
  • 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
  • DanJS
at that x=2, the function does not exist, but the , and changes increasing to decreasing
DanJS
  • DanJS
it is a hole in the graph, or a cusp, piecewise defined, i think
anonymous
  • anonymous
so its concave down at -4 and 4? because f''(c)>0?
Rizags
  • 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
  • DanJS
|dw:1449628123629:dw|
anonymous
  • anonymous
so its not concave down at ant points?
Rizags
  • Rizags
nope, its concave up on (-inf, 2) and (2, inf)
Rizags
  • Rizags
we can't know what occurs at 2 because g isn't differentiable there; its a cusp
anonymous
  • anonymous
so question 3 is a trick question?
DanJS
  • DanJS
yes, like the absolute value piecwise dfined functnon
DanJS
  • DanJS
hole in the graph rather, because here g(2) is not defined in the function
anonymous
  • anonymous
alright so what about 4 and 5? I have no idea what they are even asking about
Rizags
  • Rizags
Evaluate g'(3) to begin. That will be the slope of the tangent at x=3.
DanJS
  • DanJS
this is just a special case, the continuous functions are straight forward
anonymous
  • anonymous
g'(3) is -7?
Rizags
  • 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
  • anonymous
y-4=-7(x-3)?
DanJS
  • 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
  • DanJS
and the derivative exists
Rizags
  • 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
  • anonymous
where did you get the 4 from in the point (3,4)
Rizags
  • Rizags
in the problem, it says g(3)=4
anonymous
  • anonymous
ohhh ok
anonymous
  • anonymous
how do i determine if it lies above or below?
Rizags
  • 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
  • anonymous
so since the entire graph is concave up, the tangent line lies below?
Rizags
  • 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
  • anonymous
so i should say: at x=3, the graph is concave up, therefore the tangent line lies below it
Rizags
  • Rizags
correct.
anonymous
  • anonymous
alright thanks so much guys
Rizags
  • Rizags
ya np
DanJS
  • DanJS
welcome

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