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The function f(x)=2x^3−27x^2+48x+1 has one local minimumand one local maximum.This function has a local minimum at x equals__________with value ____________...

Question

The function f(x)=2x^3−27x^2+48x+1 has one local minimumand one local maximum.This function has a local minimum at x equals__________with value _________________and a local maximum at x equals __________________with value ______________________

The function f(x)=2x^3−27x^2+48x+1 has one local minimum and one local maximum. This function has a local minimum at x equals __________ with value _________________ and a local maximum at x equals __________________ with value ______________________



Answers

What are local maximum and minimum values of a function?

Okay, So if we have applied at seeing for are greater than or equal to off of X for all ex contained in an open interval. I This is an open interval. Ah, it contains See and FFC is a local next Ah us. And now if we have f of C is less than or equal to affect for us containing iis an opening interval then we know that FFC is a local men.

Looks like it's getting easier again. Um Oh wait a minute. Were given the derivative? Well that's actually even easier if you're given the derivative, all you have to do is set it equal to zero, then either X equals zero or one minus X equals zero and therefore X would be one done.

Mhm. This time we're given the derivative. And so we need to start by just setting it equal to zero. So we set that equal to zero and we get X. Equals zero. X. Equals one. Now we need to take the derivative second derivative to see if these are maximum or minimum. So driven to the first times the second plus the derivative of the second times the first. Just simplifying. Although I really don't need to simplify. Let's just do the second derivative of 03 times zero. That's zero, second one's time zero. Also the second derivative of zero zero E. Second derivative of one also is going to be zero also zero. Mhm. Yeah two. So I don't necessarily know whether these are maxima or minima or possibly points of inflection. So I uh did something a little more advanced. I actually did the integral. Um And then plotted what it would have looked like. Um And so the actual function would look something like this. Um Yeah so at zero and one it actually is a minima. So these are both minima. Um We can't really tell based on the data uh what the vertical translation would be, so that graph could be up or down uh could be translated or shifted up or down. Um So those two are minimum.

Now we've got this derivative and we also have an interval. Oh no, we're saying that not an interval is just saying that B is greater than A and A is greater than zero. Okay, well first of all critical points, we'll be at zero A and be drew to the first times the 2nd and 3rd plus the derivative of the second times the first times the third plus the derivative of the third times the first times the second. Okay. Zero. The 2nd and 3rd term both become 00 minus A would be a negative number and zero minus B would be a negative number. So we've got a negative number times negative number which would be a positive number which is going to make this a minimum. Mhm. A the first and the last terms would be zero. Um A is a positive number. I'm looking at the second term, A is a positive number. So I'm looking right there and then A minus B is going to be a negative number. That's going to be a max. Okay. Only the last term will be non zero. It's going to be be with a positive number times B minus A, which is also a positive number, which is gonna make it a minimum. Thank you for watching. Okay.


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