Use the Second Derivative Test to classify the relative extrema, for (x^3)+(-6√x)-3 .if the test applies, otherwise use the First Derivative Test.

Question

Hi I have been having trouble with this question because when I try to calculate the second derivative extrema i get imaginary numbers and I dont understand if those count as greater or less than 0.

The text also says that if the test does not apply, otherwise use the First Derivative Test so maybe it is that? I tried doing another question similar to this using the First derivative test for the extrema and I got the question wrong.

I could have just messed up on my calculations but I double checked on multiple calculators.

Can someone please show their work so I understand and can finally solve this?

Answer 1

I think your function is:

f(x) = x³ - 6√x - 3

or:

f(x) = x³ - 6x^1/2 - 3

So, using the power rule:

f '(x) = 3x² - 3x^-1/2

Setting f '(x) = 0:

0 = 3(x² - x^-1/2)

x² - x^-1/2 = 0

x² - 1/x^1/2 = 0

(x^1/2/x^1/2)x² - 1/x^1/2 = 0 [getting a common denominator to subtract the fractions]

x^5/2/x^1/2 - 1/x^1/2 = 0

(x^5/2 - 1)/x^1/2 = 0

The denominator cannot contribute to the expression being equal to zero, therefore:

x^5/2 - 1 = 0

x^5/2 = 1

x = 1 (the critical point)

f ''(x) = (3x² - 3x^-1/2)'

f ''(x) = 6x + 3/2x^-3/2

To use the second derivative test, we need to evaluate f ''(1). If f ''(1) > 0 then x = 1 is a local minimum and if f ''(1) < 0 then x = 1 is a local maximum.

f ''(1) = 6(1) + 3/2(1)^-3/2 = 6 + 3/2 = 7.5 (a positive number) therefore x = 1 is a local minimum. To ensure the function is concave up on both sides of the critical point, pick a number slightly lower than 1 (like 0.9) and one slightly larger (like 1.1) and check the second derivative at those points to ensure they are positive as well.

Answer 2

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