How many tens are in four hundred thirty
1 answer:
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Answer:
Step-by-step explanation:
The max and min values exist where the derivative of the function is equal to 0. So we find the derivative:
Setting this equal to 0 and solving for x gives you the 2 values
x = .352 and -3.464
Now we need to find where the function is increasing and decreasing. I teach ,my students to make a table. The interval "starts" at negative infinity and goes up to positive infinity. So the intervals are
-∞ < x < -3.464 -3.464 < x < .352 .352 < x < ∞
Now choose any value within the interval and evaluate the derivative at that value. I chose -5 for the first test number, 0 for the second, and 1 for the third. Evaluating the derivative at -5 gives you a positive number, so the function is increasing from negative infinity to -3.464. Evaluating the derivative at 0 gives you a negative number, so the function is decreasing from -3.464 to .352. Evaluating the derivative at 1 gives you a positive number, so the function is increasing from .352 to positive infinity. That means that there is a min at the x value of .352. I guess we could round that to the tenths place and use .4 as our x value. Plug .4 into the function to get the y value at the min point.
f(.4) = -48.0
So the relative min of the function is located at (.4, -48.0)