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anonymous

  • one year ago

The graph of f ′ (x), the derivative of f of x, is continuous for all x and consists of five line segments as shown below. Given f (0) = 7, find the absolute minimum value of f (x) over the interval [–3, 0].

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    First, I'm not sure but if maybe this makes sense to you, it's worth a try Ok to find the minimum value of f(x) in the interval [-3,0] we have to evaluate f(x) at the points -3, 0 and we also have to find the value x for which f'(x)=0 because the value of x for which f'(x)=0 is the value of x at which the function f(x) has a critical point(it's either maximum or minimum) Now from the graph we know f'(x)=0 for x=0 therefore at the value x=0 we have \[f(0)=7\] Since x=0 is also the end point in the interval, that's 2 out of 3 points done Now we need to look at x=-3 Consider the definite integral \[\int\limits_{-3}^{0}f'(x)dx=f(0)-f(-3)\]\[\implies f(-3)=f(0)-\int\limits_{3}^{0}f'(x)dx\] Using the graph you can find the f'(x) in the interval -3 to 0 as an equation of a straight line

  3. anonymous
    • one year ago
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    okay sorry had to go to the restroom here are the answer choices by the way 0 2.5 4.5 11.5

  4. anonymous
    • one year ago
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    okay so if the equation is a straight line then their is no minimum right ?

  5. anonymous
    • one year ago
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    Since x=0 is the value of x for which f'(x)=0, f(0) gives the value of f(x) at which it is either maximum or minimum thus the value 7 is either it's maximum or minimum now to check whether it's maximum or minimum we'll have to look at how f'(x) changes value when it passes through 0 If f'(x) changes from negative to positive ie, f'(x)>0 for some x<c and f'(x)<0 for some x>c, then it's a point of local maxima here the c is 0, it's the value of x at which f'(x) is 0 so take any value less than 0 and more than 0 and look at f'(x) and if it's f'(x)<0 and f'(x)>0 for x<c and x>c respectively then it's minima

  6. anonymous
    • one year ago
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    umm wow okay do you mean positive and negative with respect the the x or the y axis?

  7. anonymous
    • one year ago
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    take value of x less than 0, look at the corresponding value of f'(x) if it's less than 0 and take value of x more than 0 look at the corresponding value of f'(x) if it's more than 0 it's a local minima and the value of 7 would be minimum if that's the case otherwise maximum

  8. anonymous
    • one year ago
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    and if it does not change sign it's not a maxima or minima

  9. anonymous
    • one year ago
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    lol okay im lost

  10. phi
    • one year ago
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    The integral of f'(x) is the area under the curve. if we had the value at f(-4), then we could find f(x) for any x from -4 to +5 by adding the area from -4 up to the x of interest to f(-4). in this case we have f(0), so we subtract the area under the curve between x=-3 and x=0 from f(0)

  11. anonymous
    • one year ago
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    give me a second to catch my breath

  12. phi
    • one year ago
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    hopefully it is clear that starting at f(-4) the area under the curve is always going up as we move to the right... f(x) is constantly going up. if we want the min in the range of -3 to 0 then we want f(-3) ...

  13. phi
    • one year ago
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    we can say f(-3) + area under the curve from -3 to 0 = f(0) = 7 thus f(-3)= 7 - area of triangle

  14. anonymous
    • one year ago
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    if the area is increasing and then starts to decrease after x = -3 wouldn't that make x = -3 a maximum ?

  15. mathmate
    • one year ago
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    |dw:1439293299504:dw|

  16. anonymous
    • one year ago
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    Yep it seems like the point f(0) is not of our use, we must evaluate f(-3) use the equation \[f'(x)=-x\] and integrate it within the limits -3,0 because f'(x) is not changing sign for a value of x less 0 and more than 0 (it's positive in both case) so it's not a point of min or max the equation I gave is the equation of the 2nd line segment from left in the graph

  17. phi
    • one year ago
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    the area of the triangle has base 3 and height 3 , so area 4.5 and f(-3)= 7-4.5= 2.5

  18. anonymous
    • one year ago
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    where did you get f'(x) = -x

  19. phi
    • one year ago
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    ***if the area is increasing and then starts to decrease after x = -3**** the area does not decrease... as you move to the left you have more area (imagine painting the region between the curve and the x-axis) it is true the area increases at a slower rate, but it still goes up. for the area to decrease the curve must go *below the x-axis* (that counts as negative area)

  20. phi
    • one year ago
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    **as you move from left to right

  21. anonymous
    • one year ago
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    okay i see

  22. anonymous
    • one year ago
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    \[y=(\frac{y_{2}-y_{1}}{x_{2}-x_{1}})x+C\] Equation of a straight line if the line passes through origin, c=0 \[y=f'(x)=(\frac{y_{2}-y_{1}}{x_{2}-x_{1}})x\] take any pair of points lying in the line segment now \[f'(x)=\frac{2-1}{-2-(-1)}.x=\frac{1}{-1}.x=-x\]

  23. anonymous
    • one year ago
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    with this you can find f(-3) \[f(-3)=f(0)-\int\limits_{-3}^{0}f'(x)dx=7-\int\limits_{-3}^{0}-x.dx=7+\int\limits_{-3}^{0}x.dx\]

  24. anonymous
    • one year ago
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    compare the value of \[f(0)\] and \[f(-3)\] you can tell which is less, whichever value is less is the minimum value forget some of the earlier stuff I said about looking at the sign, that's not required here

  25. anonymous
    • one year ago
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    well obviously f(-3)

  26. anonymous
    • one year ago
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    okay i think i'm still unsure how to do this ,iv got another problem similar to this one could you help me with that one too?

  27. anonymous
    • one year ago
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    sure

  28. anonymous
    • one year ago
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    okay one sec

  29. anonymous
    • one year ago
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    did you understand how I got f'(x)=-x

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