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Numerade Educator

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Problem 27 Easy Difficulty

Sketch the graph of $ f $ by hand and use your sketch to find the absolute and local maximum and minimum values of $ f $. (Use the graphs and transformations of Section 1.2 and 1.3).

$ f(x) = \left\{
\begin{array}{ll}
x^2 & \mbox{ if}-1 \leqslant x \leqslant 0\\
2 - 3x & \mbox{ if} 0 < x \leqslant 1
\end{array} \right. $

Answer

No absolute or local maximum.
Absolute minimum $f(1)=-1$. Local minimum $f(0)=0$.

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Video Transcript

we're going to sketch the graph of the function defined by pots like this. X square if X is Rather than or equal to 91 and less than or equal to zero and two X minus three to minus three X. If X is greater than zero and less than or equal to one. There is a piecewise function with two parts. One for the interval negative 10 including both negative 10 and the other From 0 to 1 excluding zero. But including one. The first part of Super Public Square. The second person straight line. Too many three eggs. And when we do that graph the hand, we would use it to find the absolute and local maximum and minimum values of the function. So the main of these functions we see here is negative 11. Close interval, we have two parts so we have to draw each part of the function, the first part of the parable like square, but between negative one and Syria including -1 and including Syria. So we have this coordinate system here at negative one, X square is one. So we get this point of the problem is included in the graph and at zero we get x square is equal to zero. So we had these point here, in fact I have forgotten to put here this Y axis. Okay, so we have two points and we know the problem is something like this. Without this seen here. But that's the idea. So that the problem, the other parties a straight line, we can work with some transformation of the identity. But because this is the line we know sufficient to try to points and join them. So in the in the case zero were included in the formula will will have to but it's not included but we know the value should be too. So we have this, we draw this circle Without field to indicate. The point is not included in the graph. And then we have the straight line between that point and another point we can choose is when x equal one we get to minus three times one is two minus three which is negative one. So at one the function has this value -1. And these values included in the graph. So I indicated with his I feel red a circle So we are ready to draw the line. We are going to use this here and this line is more or less this step. Yeah can tune it here now. So we have this point connected with this other point. So we get to rotate a little bit and we are almost there and then we Okay, a little bit mm let's see like this. Okay we are almost here let's see something like that. Mhm Okay here we are. Yeah. So this is the graph of the function by Yeah. This way it functions with the parts two pieces And we can say this this continuous function at zero for example. But let's see the properties about the extreme. 1st. We can see we have an absolute minimum here at this point, Dizzy should always point in the graph is included in the in the graphic as is the image of one Here which uses this formula here and the value because funding body is -1. So we can say that F has an absolute minimum value. The value is -1. And that value of course at x equal *** last one. Now for the local minimum absolute minimum and for the local minimum in fact there is one here. You can say that this value here, zero at zero is a local minimum because if we take any interval or at least this interval I drawn here around zero. All the images are greater than the images zero which is zero, that is all the images that are positive. We can see some of them are on the line and the others on the problem but all those images are greater than or equal to the image of serum. So zero is local minimum. So F has a local and is the only one with his property in the draft so has all local minimum. Mhm. Value zero, add x equals suit. Mhm Okay, respect to the maximum value, there is no absolute maximum because the highest fine should be at this point but it's not included in the graph. So the function is always taking greatest values closer and closer to the 0.0.20 or 02. Sorry, but it's not, it never reaches that values of there is no absolute maximum. Have has no absolute maximum respect to local maximum. Uh There there's there's no local maximum either because we have in all the points, we always have taken any interval around the point. We always have images that are greater or lower than the point. So there's there are no local maximum, so no absolute or local maximum. But yes, so this is the These are properties of dysfunction defined by two pieces.