Problem 40 Medium Difficulty

(a) Find the intervals of increase or decrease.
(b) Find the local maximum and minimum values.
(c) Find the intervals of concavity and the inflection points.
(d) Use the information from parts $ (a) - (c) $ to sketch the graph.
Check your work with a graphing device if you have one.

$ g(x) = 200 + 8x^3 + x^4 $

Answer

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01:55

Carson M.

Watch More Solved Questions in Chapter 4

Video Transcript

Yeah, In this problem, we are learning how the derivative effects the shape of a graph. So specifically, we're going to be using the 1st and 2nd derivative test to determine things like where our function is increasing or decreasing and the con cavity of our function. Now you might be asking, Why do I need to know this? Well, this gives us a lot of information about the functions behavior, but also this is going to come in handy when you're learning how to curve sketch in calculus. This is all leading up to that. So for part A were given the function G of X equals 200 plus eight x cubed plus X to the fourth. Now we need to find the intervals of increase or decrease. So the first thing we should do is take the derivative we'll take G Prime of X will get 24 x squared plus four x cubed. And then we can factor out a term that is common with both of the terms in our derivative so we can factor out of four x squared. So when we do that, we get four x squared times six plus X and then we need the critical numbers of the function. So we'll set this simplified derivative equal to zero and solve for X well X zero on X is negative. Six. So what we see is you can plug in a point into our original function and determine if it's increasing or decreasing. And if you do that, you can see that ffx is increasing when X is greater than negative six. And ffx is decreasing when X is less than negative. Six. I'm sorry. This should be G of X. It's still the function. Um, my apologies now for Part B, we're told. Where is the local maximums or minimums? Well, we have a local minimum by the first derivative test at X equals negative six so we can plug in negative six and to our original function, G. And we'll get the other coordinate. So if we plug in, negative six will take G of negative six and I'll get negative to 32. So that is the point that we see a local minimum in our graph now for part C, we're told, Well, let's determine the con cavity of our function. Now when you hear the word con cavity. You should know that this is directly associated with the second derivative. You can't determine Con cavity with the first derivative. We need the second so we'll take G double prime of X and we'll get four X. Pardon me 48 X plus 12 x squared again. Factor out a term to make it easier to find. The critical numbers will factor out a 12 X and then we'll multiplied by four plus X well cited equal to zero solve for X X zero and X is negative four. So now we could makes minerals and determined con cavity. So our first interval would be negative infinity to negative for. And then you can plug in a point within that interval and determine the sign of the function. And when you do, we determine that this is con cave up in the interval and then our second interval is negative for 20 and that would be con cave down. And then finally we have zero to infinity and that would be Khan gave up. So what we can see here is he went from concave up to down, toe up. That means we have an inflection point, we have an inflection point in X equals negative four and X equals zero. And then finally, we're told, Well, what's going to happen if we graph this? Will the information we found match the graph of the function and it does. You can see the graph. We see the same intervals of increasing decreasing con cavity. And so this is a great way to check your work. Um, or you can work backward from the graph also. So I hope that this problem helped to understand a little bit more about how the derivative affect the shape of the graph and how we can use differentiation to find things like the interval of increase in decrease and the con cavity.