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Does it have to do with the the slope?
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Do we all agree the answer is C? The hard part is ruling out D. But if we look at the interval \[0.5, 1.5\] in D, we see the function goes over 1 and up 2, so the derivative (h(x)) in the top picture would need to go as high as 2. Since the derivative h(x) only goes up to 1.5, we can rule out D.
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You might practice this kind of analysis on the geometric calculator the geogebra offers. May start with quadratic functions.
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The easiest method is to match up the x coordinate of the zero crossings in the derivative with the local extrema (peaks and troughs) in the graph below.  A derivative going from negative to positive will be a trough, a derivative going from positive to negative will be a peak.  That narrows it down to two.  Then look at how high or low the derivative gets.  It goes past 1 and -1.  That means the slope of the graph below will have steeper slopes than a diagonal (which would be 1 or -1).  That only leaves D.
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At a stationary point the derivative graph crosses the x-axis.
When the function is decreasing the graph is below the x axis.
When the function is increasing it's above the x-axis.
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The point on the graph of the derivative is the slope.

So at points where you’re going from positive slope to negative slope, or vice versa, (if you know the graph is continuous and you can find it’s derivative) is where your graph of the derivative will hit the x axis.

Only one of these graphs match have zeroes where there’s zero slope.
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Don’t ask about “rules”, cuz you’re gonna get really confused. You need to think about what the derivative actually is first. And yes, the derivative function has everything to do with the slope of the original function.

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