Question

How to find the relative extrema of a function? Lets say x^2+2x

Answer 1

for that one all you need is that
\[y=x^2+2x\] is a parabola that opens up, so it will have no maximum, but it will have a minimum value
the minimum value will be the second coordinate of the vertex.
first coordinate of the vertex is
\[-\frac{b}{2a}=-\frac{2}{2}=-1\] in this example.
second coordinate is what you get when you replace x by -1 , namely
\[y=(-1)^2+2\times (-1)=-1\]

Answer 2

therefore -1 is your relative and absolute minimum, there are no other extrema

Answer 3

So this is what I will need to do for every question? -b/2a then plug that # in for x?

The question on my review sheet is f(x) = sinh(x)-(x-1)cosh(x) how would we go about doing that?

I was actually thinking about the whole -b/2a and that has to do with finding the bounds of a volume to rotate around a solid correct? I also learned that you can set f(x) and g(x) = and find the bounds that way... Diff question, but yeah...

Answer 4

this is not what you need for every question. this is only what you need if you have quadratic

Answer 5

\[f(x) = \sinh(x)-(x-1)\cosh(x)\] is quite a different story. for this you need to take the derivative, find the crictical points, and then see whether they correspond to maxima or minima

Answer 6

critical points are the points that are interersected or...? Also Idk how to calculate the max or min... :( I have 0 knowledge about relative stuff...

Answer 7

critical points are where the function exists at the point, but the derivative equals 0 or doesn't exist for that point

Answer 8

so we should start out by saying e^x - e^-x/2 - (x-1) * (e^x+ e^-x/2)? Then plug in for 0?

Answer 9

you would find the derivative of the function and set the entire derivative equal to 0 to find the x-values where it is 0.

Answer 10

so, if we found the derivative of this function..
f(x)=sinh(x) - (x-1)cosh(x) (it may help to distribute the -1 into that x-1 to help in getting the derivative)
=sinh(x) + (-x+1)cosh(x)
=sinh(x) + (1-x)cosh(x)

f'(x) = cosh(x) + (-1)(cosh(x)) + (1-x)(sinh(x))
= cosh(x) - cosh(x) + (1-x)(sinh(x))
= (1-x)(sinh(x))

sinh(x) = 0 and 1 - x = 0
x = 0 1 = x

Answer 11

determining whether it is a maximum or minimum means you would check the derivative of points around your critical points in order to determine the signs

since the critical points are x=0 and x=1, you'd test points in these intervals in the derivative function:
x < 0, 0 < x < 1, and 1 < x

if the derivative changes between two points, then there is an extremum there -- if it is from positive to negative, it is a maximum; and if it is from negative to positive, then it is a minimum

Answer 12

ah.....

Answer 13

what exactly do you mean by "if the derivative changes between 2 points?"

Answer 14

between a point in one interval and a point in the next, would probably be a better way to word that...


this is basically what we're doing with the function and its derivative -- we find one derivative that is to the left of a critical point, and one to the right of it (both not being beyond any other critical points) to see if the graph does something like that around that point