The figure below shows the graph of f ′, the derivative of the function f, on the closed interval from x = −2 to x = 6. The graph of the derivative has horizontal tangent lines at x = 2 and x = 4.
Find the x-value where f attains its absolute maximum value on the closed interval from x = −2 to x = 6. Justify your answer

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- ganeshie8

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- anonymous

|dw:1439314520378:dw|

- ganeshie8

what do you know about the relationship between first derivative and relative min/max ?

- anonymous

wherever the first derivative = zero their is likely a max min , where the first derivative is negative and then positive the function is concave down and the inverse of that to indicate concave up

- anonymous

umm thats pretty much it, am i missing something

- ganeshie8

that looks good, look at the given graph of first derivative
notice that the first derivative stays negative in the interval [-2, 5],
this means the function is "decreasing" in this interval, yes ?

- anonymous

yep

- ganeshie8

we cannot have any max/min in the interval (-2, 5) since the function is continuously decreasing

- ganeshie8

what about the point (5, 0)
does that mean the function has a min or max at x=5 ?

- anonymous

min

- ganeshie8

good, since the first derivative is going form "negative" to "positive", the function will have a local minimum at x=5

- ganeshie8

that essentially means, we do not have any local maximums in the interval (-2, 6)

- ganeshie8

so the absolute maximum must occur at the boundary points

- anonymous

x = 6!!

- ganeshie8

how do you know ?

- anonymous

well it just kept increasing after x = 4 so i figured it would be the highest , but i guess x=-2 could be just as high because we don't know what happened before x = -2 right?

- ganeshie8

Notice that the first derivative is "negative" in the interval (-2, 5)
that means the actual function is "decreasing" in the interval (-2, 5)

- anonymous

okay i see so your suggesting that because it decreased for such a long interval and then increased for such a short interval the maximum would be ant x = -2
is that what your saying?

- ganeshie8

Kindof! but thats not it, at this point, i do believe the function attains its maximum at x = -2

- anonymous

okay well why do you suggest its x = -2?

- ganeshie8

|dw:1439315773571:dw|

- ganeshie8

which area do you think is more
black or red ?

- anonymous

well grey obviously

- ganeshie8

That means the actual function did not recover the fall yet

- ganeshie8

so x=-2 is indeed the absolute maximum

- anonymous

okay that makes sense

- ganeshie8

If you're good with definite integrals, notice below :
\[f(6)-f(-2) = \int\limits_{-2}^6f'(x)\,dx \lt 0\]
that implies \(f(6)\lt f(-2)\)

- anonymous

okay thanks for the help!

- ganeshie8

yw

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