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].

- anonymous

- chestercat

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- anonymous

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- 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

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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## More answers

- anonymous

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

- 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

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

- 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

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

- anonymous

lol okay im lost

- 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

give me a second to catch my breath

- 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

we can say
f(-3) + area under the curve from -3 to 0 = f(0) = 7
thus f(-3)= 7 - area of triangle

- anonymous

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

- mathmate

|dw:1439293299504:dw|

- 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

the area of the triangle has base 3 and height 3 , so area 4.5
and
f(-3)= 7-4.5= 2.5

- anonymous

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

- 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

**as you move from left to right

- anonymous

okay i see

- 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

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

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

well obviously f(-3)

- 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

sure

- anonymous

okay one sec

- anonymous

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

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