## 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].

1. anonymous

2. 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

3. anonymous

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

okay so if the equation is a straight line then their is no minimum right ?

5. 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 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

umm wow okay do you mean positive and negative with respect the the x or the y axis?

7. 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

8. anonymous

and if it does not change sign it's not a maxima or minima

9. anonymous

lol okay im lost

10. 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)

11. anonymous

give me a second to catch my breath

12. 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) ...

13. phi

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

if the area is increasing and then starts to decrease after x = -3 wouldn't that make x = -3 a maximum ?

15. mathmate

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16. 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

17. phi

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

where did you get f'(x) = -x

19. 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)

20. phi

**as you move from left to right

21. anonymous

okay i see

22. 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 $f'(x)=\frac{2-1}{-2-(-1)}.x=\frac{1}{-1}.x=-x$

23. anonymous

with this you can find f(-3) $f(-3)=f(0)-\int\limits_{-3}^{0}f'(x)dx=7-\int\limits_{-3}^{0}-x.dx=7+\int\limits_{-3}^{0}x.dx$

24. 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

25. anonymous

well obviously f(-3)

26. 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?

27. anonymous

sure

28. anonymous

okay one sec

29. anonymous

did you understand how I got f'(x)=-x