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Problem 54

a. Find the local extrema of each function on the…

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Problem 53

a. Find the local extrema of each function on the given interval,
and say where they occur.
b. Graph the function and its derivative together. Comment on
the behavior of $f$ in relation to the signs and values of $f^{\prime}. $
$$
f(x)=\sin 2 x, \quad 0 \leq x \leq \pi
$$

Answer

$$
=\frac{3 \pi}{4}
$$


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Video Transcript

So for a day's question for this problem, our function is ffx equaled. Sign two x In order to find local extremist, The very first thing that we need to do is to find the derivative off the given function. F dash of facts is going to be too cause two X This door abated was evaluated by using chain rule off derivatives to find the local extremists. We always equate F dash to zero in this case the value off F dash that we have found using their chain rule of Terra Vader's. As to costs two. X equals 20 When we simplify this equation, we get cause two X equals do zero. This particular solution holds true for the following values Effects pie by four and three pie by four In order to find the local extremists, we next plot these values for the given range ever given. Grange is X is less than or equal to pi, but crater in equal to zero. So we plot these points on a number line Our do mean is zero and pie. Rather, I were demeanors between the values of zero and pie. We put these evaluated values and plod them right onto our number. Lang. So pie by four is going to be approximately over here, and three pie by four would be plotted right over here. When fighting local extremists, be figure out the intervals first. In next we calculate if, for those giving intervals the value off our function f dash off at the rather, we find him. Rather, we find it intervals. And then we find if the values off F gash effects is going to be positive or negative for any given interval. In this case, over here we have three intervals oneness between zero and pie by four. Another is between pie by four and three pie by four, and the last one is between three. Pie by four and pie. For this interval, the value off costs of two FX is always going to be positive. For this second interval, the value off cost two effects is always going to be negative. The key over here is to consider any number between this interval and calculate the value of cost of two acts, for that's for that number. So say we consider pie by two. So when you put pie by two in this equation over here it is always going to be negative because costs of pie is negative. One for this interval cause of two acts is always going to be positive, said this. Right Here are your three intervals for local minimum per the rules Tates for local minimum. The values always change from negative to positive, which holds true for three pie by four. Hence, three pie by four is our local minimum. If he put three pie by four in place off X in our function, our final answer would be understood through two divided by two. As for local maximum, the rules tapes that the values change from positive to negative, which holds two True for pie by four. Because for pie before the DellaVedova is equal to zero. When you put pie by four up in your function, your value is one

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