f(x) = 3x6 - 7x5 Find all critical numbers of f . If there are no critical numbe
ID: 3212491 • Letter: F
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f(x) = 3x6 - 7x5 Find all critical numbers of f . If there are no critical numbers, enter 'NONE'. Critical numbers = Use interval notation to indicate where f (x ) is increasing. Note: Use INF for infinitive, '-INF for - infinitive. and use 'U' for the union symbol. Increasing: Use interval notation to indicate where f ( x ) is decreasing. Decreasing: Find the x-coordinates of all local maxima of f . If there are no local maxima, enter 'NONE'. x values of local maxima = Find the x-coordinates of all local minima of f . If there are no local minima, enter 'NONE'. x values of local minima = Use interval notation to indicate where f ( x ) is concave up. Concave up: Use interval notation to indicate where f ( x ) is concave down. Concave down: List the x values all inflection points of f . If there are no inflection points, enter 'NONE'. x value of inflection points =Explanation / Answer
The critical points are where the first derivative equals zero, as you've already calculated.
The places where it is increasing can be found by testing the ORIGINAL function around these points. Pick values such that you encompass the two critical points and find the value of the function at those points. (I'd pick -1, 1, and 3 for example).
You'll find that the function will behave as follows: (-) @x=-1, (-) @x=1, and (+) @x=3
This means that the VALUE of the function has those signs at those points, NOT that it is either increasing or decreasing there.
Because the function is negative from the left until x=0, and then negative afterwards, it must increase until zero, and then decrease afterwards. We know this because of the definition of a critical point (Function value is either not changing or undefined). It decreases until x=35/18, where it has another critical point. A distance after this point, the function is again positive, so it increases indefinitely because there are no more critical points.
If we had taken a value of 2 instead of 3, we could have gotten a value of the function as still negative, but the value would have been greater than the value at the critical point. Thus, it may not always be necessary to compare the actual values to find the trends of a graph, but it IS always good practice.
B: (-INF,0], [(35/18),INF) C: [0,(35/18)]
To get the local maxes and mins, simply observe the values you computed before and find what happens to the function around the critical points. If it falls on either side, it is a local max. If it rises on either side, it is a min.
D: x=0 E: x=35/18
Concavity and inflection are characteristics of the SECOND derivative of a function. Wherever the value of the second derivative is positive, it is concave up. Wherever it is negative, it is concave down. Wherever it is zero, the function has an inflection point.
f''(x)=90x4-140x3. So if we set this to zero, we find inflection points at x=0 and x=14/9. The same test is used for finding the concavity as the test for increasing/decreasing. Take test points (I used the same 3 values), and find that it is concave up from negative infinity until x=14/9, and concave down from there till positive infinity.
F: (-INF,(14/9)] G: [(14/9),INF) H: x=14/9
I hope this helps, feel free to ask questions!
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