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
@Empty @IrishBoy123
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what do you know about the relationship between first derivative and relative min/max ?
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
umm thats pretty much it, am i missing something
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 ?
yep
we cannot have any max/min in the interval (-2, 5) since the function is continuously decreasing
what about the point (5, 0) does that mean the function has a min or max at x=5 ?
min
good, since the first derivative is going form "negative" to "positive", the function will have a local minimum at x=5
that essentially means, we do not have any local maximums in the interval (-2, 6)
so the absolute maximum must occur at the boundary points
x = 6!!
how do you know ?
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?
Notice that the first derivative is "negative" in the interval (-2, 5) that means the actual function is "decreasing" in the interval (-2, 5)
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?
Kindof! but thats not it, at this point, i do believe the function attains its maximum at x = -2
okay well why do you suggest its x = -2?
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