Question

Find the relative minimum of

Answer 1

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:

$y'=9x^2+28x-11$

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)