Sketch the graph of a continuous function ƒ on [0, 4] satisfyi...

Question

Sketch the graph of a continuous function ƒ on [0, 4] satisfying the given properties.

ƒ' (x) and ƒ'3 are undefined; ƒ'(2) = 0; has a local maximum at x= 1; ƒ has local minimum at x = 2; and ƒ has an absolute maximum at x= 3; and ƒ has an absolute minimum at x = 4 .

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Do the following properties correctly describe the graph of FF X on square bracket 0.8 square bracket, justify if incorrect. F2 and F4 are undefined. F 6 is equal to 0. F of X has local minima at X equals 4, and X equals 6. F of X has a local maximum at X equals 2. F of X has an absolute maximum at X equals 0. FF X has an absolute minimum at X equals 6. Awesome. So it appears for this particular problem, we're asked to determine whether or not the following properties that are given to us by the problem itself correctly describe the graph that is given to us of F of X on this closed interval of 0.8. That's what the square brackets mean. The square brackets indicate that this is a closed interval, and we're asked to justify if these properties are incorrect, if they are incorrect. So before we start solving, let's investigate our graph. So as it appears, we have our Y axis, which is denoting our vertical axis, and then our horizontal axis is the X axis, and we have a blue curve for F of X. And this blue curve almost looks like it's shaped like a number 3 that's like been flipped upside down and it's diagonal. It has a jagged edge and a curved edge that's parabolic, but then it also has a jagged edge that resembles a V graph. So, and it appears it starts at on it touches the Y axis at 08 And it also touches the X axis at So it's so it's actually it's gonna be at 8, sorry, 80 for the Y axis and then at 06 for the X axis. If you go up 0 and then over 6. Whereas on the Y axis you go up. 8 and over 0. So those, so those are the two points where it touches the X or Y axis, and it appears that it ends at 8. 4 at the corner 0.84. Fantastic. So now that we have a good visualization of what's happening for this particular problem, let us, in order to solve for this problem, first off, let us examine each property that is provided to us based on the given graph. So we need to note that F Of 2 and F of 4 are undefined, so this is correct. So, let's make a note of this. So this will be equal to undefined. So I said UD for undefined. So that means that this is correct, so I'll give it a red check mark to indicate that this is correct. Awesome. So that means that this specific property that says that F2 and F4 are undefined, so that is correct, because there is a presence of a sharp turn at X equals 2, like we were discussing earlier, and X equals 4 suggests that F of X is undefined at these points. We can also state that F of 6 is equal to 0. We could say that this statement is correct because the function smoothly transitions at X equals 6, reaching its lowest point, and this suggests that there's a horizontal tangent. Therefore, F 6 will be equal to 0. And F of X has local minima at X equals 4 and X equals 6. However, this particular piece of information for the property is incorrect. So F of X has a local minima, so L Min, so L min, let's make a note of that. So L and also the hyphen, we'll call it like quotation marks, L hypha min. So it has a min at X equals 4 so. So this whole information for L min, so this local minima at X equals 4 and X equals 6 will be incorrect. I'll put a red X. So this is incorrect because at X equals 4, the function transitions from increasing to decreasing. So therefore, it is a local maxima, and at X equals 6, the function transitions from a decreasing situation to an increasing situation. So because it transitions from decreasing to increasing, therefore it is a local minima. Or minimum, I should say. And the statement that states that F of X has a. Local maximum. And we'll do L hyphen max so local maxima at X equals 2. This piece of information is incorrect, and it's incorrect because, and we'll put a red X. So once again, F of X has a local maximum at X equals 2, and this is incorrect because at X equals 2, the function transitions from a decreasing state to increasing state. So therefore it is a local minimum. The statement that states that F of X, so now looking at our different property states, so it says that F of X has an absolute maximum. So we'll do it A hyphen max, so an absolute maximum. At X equals 0 is correct. So F of X is an absolute maximum at X equals 0 because the highest point on the graph is X equals 0, which makes it an absolute maximum. And finally examining our last property, which states that F of X has an absolute minimum. At X equals 6, and this is correct. So there is an absolute minimum at X equals 6 because the lowest point on the graph is X equals 6, making it an absolute minimum. So therefore, without a doubt, the given properties are partially incorrect since FFX has a local minimum at X equals 2 and a local maximum at X equals 4. So therefore, in conclusion, we can state that our final answers as a summary has to state that no, F of X has a local minimum at X equals 2, and a local maximum at X equals 4. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Sketch the graph of a continuous function ƒ on [0, 4] satisfying the given properties.

ƒ' (x) and ƒ'3 are undefined; ƒ'(2) = 0; has a local maximum at x= 1; ƒ has local minimum at x = 2; and ƒ has an absolute maximum at x= 3; and ƒ has an absolute minimum at x = 4 .

Key Concepts

Derivative and Critical Points

Local and Absolute Extrema

Continuity of Functions

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