anonymous
  • anonymous
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].
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
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
anonymous
  • anonymous
okay sorry had to go to the restroom here are the answer choices by the way 0 2.5 4.5 11.5

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anonymous
  • anonymous
okay so if the equation is a straight line then their is no minimum right ?
anonymous
  • anonymous
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 xc, 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 xc respectively then it's minima
anonymous
  • anonymous
umm wow okay do you mean positive and negative with respect the the x or the y axis?
anonymous
  • anonymous
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
anonymous
  • anonymous
and if it does not change sign it's not a maxima or minima
anonymous
  • anonymous
lol okay im lost
phi
  • phi
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)
anonymous
  • anonymous
give me a second to catch my breath
phi
  • phi
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) ...
phi
  • phi
we can say f(-3) + area under the curve from -3 to 0 = f(0) = 7 thus f(-3)= 7 - area of triangle
anonymous
  • anonymous
if the area is increasing and then starts to decrease after x = -3 wouldn't that make x = -3 a maximum ?
mathmate
  • mathmate
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anonymous
  • anonymous
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
phi
  • phi
the area of the triangle has base 3 and height 3 , so area 4.5 and f(-3)= 7-4.5= 2.5
anonymous
  • anonymous
where did you get f'(x) = -x
phi
  • phi
***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)
phi
  • phi
**as you move from left to right
anonymous
  • anonymous
okay i see
anonymous
  • anonymous
\[y=(\frac{y_{2}-y_{1}}{x_{2}-x_{1}})x+C\] 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 \[f'(x)=\frac{2-1}{-2-(-1)}.x=\frac{1}{-1}.x=-x\]
anonymous
  • anonymous
with this you can find f(-3) \[f(-3)=f(0)-\int\limits_{-3}^{0}f'(x)dx=7-\int\limits_{-3}^{0}-x.dx=7+\int\limits_{-3}^{0}x.dx\]
anonymous
  • anonymous
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
anonymous
  • anonymous
well obviously f(-3)
anonymous
  • anonymous
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?
anonymous
  • anonymous
sure
anonymous
  • anonymous
okay one sec
anonymous
  • anonymous
did you understand how I got f'(x)=-x

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