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erik [133]
3 years ago
11

Given f'(x)=(x-4)(6-2x) find the x-coordinate for the relative minimum of the graph f(x).

Mathematics
1 answer:
vlabodo [156]3 years ago
5 0
<span>The graph you plotted is the graph of f ' (x) and NOT f(x) itself. </span>

Draw a number line. On the number line plot x = 3 and x = 4. These values make f ' (x) equal to zero. Pick a value to the left of x = 3, say x = 0. Plug in x = 0 into the derivative function to get

f ' (x) = (x-4)(6-2x)
f ' (0) = (0-4)(6-2*0)
f ' (0) = -24

So the function is decreasing on the interval to the left of x = 3. Now plug in a value between 3 and 4, say x = 3.5

<span>f ' (x) = (x-4)(6-2x)
</span><span>f ' (3.5) = (3.5-4)(6-2*3.5)
</span>f ' (3.5) = 0.5

The function is increasing on the interval 3 < x < 4. The junction where it changes from decreasing to increasing is at x = 3. This is where the min happens.

So the final answer is C) 3
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