Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x)

Question

Consider the equation below. (If an answer does not exist, enter DNE.)

f(x) = x⁴ ln(x)

(a) Find the interval on which f is increasing. (Enter your answer using interval notation.)

Find the interval on which f is decreasing. (Enter your answer using interval notation.)

(b) Find the local minimum and maximum values of f.

local minimum value

local maximum value  

(c) Find the inflection point.

(x, y) =

Find the interval on which f is concave up. (Enter your answer using interval notation.)

Find the interval on which f is concave down. (Enter your answer using interval notation.)

Answer 1

f(x)=x⁴ln(x)

f'(x) = x⁴(1/x) + ln(x)[4x³] = x³ + 4x³ln(x) = x³[4ln(x) + 1]

f(x) has a domain x > 0, so any critical numbers must be greater than 0.

f'(x) = 0:

x³[4ln(x) + 1] = 0 when x³ = 0 or 4ln(x) + 1 = 0. x = 0 is ruled out as a critical number, so:

4ln(x) + 1 = 0

ln(x) = -1/4

x = e^-1/4 ≈ 0.7788

Applying the 1st derivative test:

f'(1/2) < 0, so f is decreasing from (0, e^-1/4].

f'(1) > 0, so f is increasing from [e^-1/4, ∞).

That means (e^-1/4, f(e^-1/4)) is a local minimum. (y ≈ -0.092)

There is no local maximum value.

f'(x) = x³[4ln(x) + 1]

f''(x) = x³(4/x) + [4ln(x) + 1]3x² = 4x² + 12x²ln(x) + 3x² = 7x² + 12x²ln(x) = x²[12ln(x) + 7]

The test for concavity means setting f'' equal zero and solving for x:

x² = 0 OR 12ln(x) + 7 = 0

x = 0 is rejected (not in the domain of f).

ln(x) = -7/12

x = e^-7/12 ≈ 0.558

Test for concavity:

f''(1/2) < 0, so f is concave downward on (0, e^-7/12).

f''(1) > 0, so f is concave upward on (e^-7/12, ∞).

Since the concavity changes at x = e^-7/12, (e^-7/12, f(e^-7/12)) is a point of inflection (y ≈ -0.057).

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