Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x⁴ ln(x)
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.)
Find the local minimum and maximum values of f. local minimum value local maximum value.
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.)

Solution:

Given, the function is f(x) = x⁴ln(x)

a) we have to find the interval in which f is increasing.

First derivative of f(x) is

f’(x) = \(\frac{d}{dx}[x^{4}ln(x)]\)

By using product rule of derivative,

[u(x).v(x)]’ = u’(x)v(x) + u(x)v’(x)

So, f’(x) = 4x³ ln(x) + x⁴ (1/x)

f’(x) = 4x³ln(x) + x³

f’(x) = x³[4ln(x) + 1]

To find critical points,

f’(x) = 0

x³[4ln(x) + 1] = 0

⇒ x³ = 0

x = 0

⇒ 4ln(x) + 1 = 0

4ln(x) = -1

ln(x) = -1/4

x = \(e^{\frac{-1}{4}}\)

x = 0.78

To find the interval where f(x) is positive or negative,

0 < x < 0.78 the function f(x) is decreasing

x > 0.78 the function is increasing

The function is decreasing in the interval (0, 0.78) and increasing in the interval (0.78, ∞).

b) we have to find the local minimum and maximum values of f.

An extremum point is found where f is defined and f’ changes signs.

Between 0 and 0.78, the function decreases at point and defined at point 0.78

After 0.78, it increases and f is defined.

So, x = 0.78 is a point of minimum and its y value is found by

f(0.78) = (0.78)⁴ln(0.78)

= (0.78)⁴l(-0.2485)

= -0.092

The point of minimum is (0.78, -0.092)

c) we have to find the point of inflection.

Second derivative of the function is

f’’(x) = \(\frac{d^{2}}{dx^{2}}[x^{3}(4ln(x)+1)]\)

f’’(x) = 3x²[4ln(x) + 1] + 4x²

f’’(x) = x²[12ln(x) + 7]

f’’(x) = 0

x²[12ln(x) + 7] = 0

⇒ x² = 0

x = 0

⇒ 12ln(x) + 7 = 0

12ln(x) = -7

ln(x) = -7/12

x = \(e^{\frac{-7}{12}}\)

x = 0.56

Put x = 0.56 in the equation

f(0.56) = (0.56)⁴ln(0.56)

= (0.56)⁴(-0.5798)

f(0.56) = -0.06

The inflection point is (0.56, -0.06)

The function is concave down when f’’(x) < 0 and concave up when f’’(x) > 0

We know the critical points are 0 and 0.56

f’’(0.1) = (0.1)²[12ln(0.1) + 7]

f’’(0.1) = -0.21, i.e. concave is down

Therefore, the function is concave down in the interval (0, 0.56)

f’’(0.7) = (0.7)²[12ln(0.7) + 7]

f’’(0.7) = +1.33 i.e. concave is up.

The function is concave up in the interval (0.56, +∞)

Therefore, 

• The interval on which f is increasing is x > 0.78
• The interval on which f is decreasing is 0 < x < 0.78
• The local minimum and maximum values of f is (0.78, -0.09)
• The inflection point. (x, y) = (0.56, -0.06)
• The interval on which f is concave up is (0.56, +∞)

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Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x⁴ln(x)

Summary:

Consider the equation f(x) = x⁴ln(x) . The interval on which f is increasing is x > 0.78. The interval on which f is decreasing is 0 < x < 0.78. The local minimum and maximum values of f is (0.78, -0.09). The inflection point. (x, y) = (0.56, -0.06). The interval on which f is concave up is (0.56, +∞). The interval on which f is concave down is (0, 0.56)