Question

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].

Answer 1

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

Answer 2

okay sorry had to go to the restroom
here are the answer choices by the way
0
2.5
4.5
11.5

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

lol okay im lost

Answer 9

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)

Answer 10

give me a second to catch my breath

Answer 11

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) ...

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

***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)

Answer 18

**as you move from left to right

Answer 19

okay i see

Answer 20

\[y=(\frac{y_{2}-y_{1}}{x_{2}-x_{1}})x+C\]
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
\[f'(x)=\frac{2-1}{-2-(-1)}.x=\frac{1}{-1}.x=-x\]

Answer 21

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

Answer 22

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

Answer 23

well obviously f(-3)

Answer 24

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?

Answer 25

sure

Answer 26

okay one sec

Answer 27

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