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anonymous

  • 5 years ago

f(x) = x4 - 32x2 + 7 (a) Find the intervals on which f is increasing. (Enter the interval that contains smaller numbers first.) ( , ) ∪ ( , ) Find the intervals on which f is decreasing. (Enter the interval that contains smaller numbers first.) ( , ) ∪ ( , ) (b) Find the local minimum and maximum values of f. (min) (max) (c) Find the inflection points. ( , ) (smaller x value) ( , ) (larger x value) Find the intervals on which f is concave up. (Enter the interval that contains smaller numbers first.) ( , ) ∪ ( , ) Find the interval on which f is concave down. ( , )

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  1. anonymous
    • 5 years ago
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    I'm going to assume you are in a form of a calc class, before I continue, correct?

  2. anonymous
    • 5 years ago
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    Yes

  3. anonymous
    • 5 years ago
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    a) You will have to find the first derivative of your equation. Once you do that, set that equation equal to 0; this will give you "critical" points (where the function switches from increasing[positive] to decreasing [negative] and from decreasing to increasing). Since part c will have similar situations (only you have to take the second derivative, or the derivative of the derivative), I will do part a: f'(x)=4x^3 - 64x 0=4x^3-64x 0=4x(x^2-16) So x=0,4, and -4. These will be your critical values. So now you need to figure out whether each interval is increasing or decreasing. Technically these will be your intervals: (-inf,-4)U(-4,0)U(0,4)U(4,inf). Let's test the first interval. So you can pick any number between -inf and -4 and put it back into your derivative; if you get a negative value, that interval is decreasing and if you get a positive value that interval is increasing. So let's try -5: then 4(-5)^3-64(-5)=-180. So the interval (-inf,-4) is decreasing. So try another number that will be in the next interval and see if it is negative/positive to determine if it is increasing or decreasing.

  4. anonymous
    • 5 years ago
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    b) Now based on whatever you get in part a, you can determine what will be your minimums and maximums. If you go from an increasing interval to a decreasing interval, you will have a maximum at the point it switches. If you go from a decreasing interval to an increasing interval, you will have a minimum at the point it switches at.

  5. anonymous
    • 5 years ago
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    c) As indicated earlier, this is very similar to what you did in part a. If you take the derivative of the derivative (f"(x)=12x^3 - 64), and set it equal to 0 you will get your "inflection points". Again you will have intervals like how you did in part a, and if numbers within that interval are negative, that interval will be considered to be concave down. If numbers within an interval are positive, that interval will be concave up. Does all of that make sense? Let me know.

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