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