f(x)=ln(x^4+27)a. Find the interval on which f is increasing and decreasing. b. Find the local minimum value of f c. Find the inflection points (2) d.Find the interval on which f is concave...

f(x)=ln(x^4+27)

a. Find the interval on which f is increasing and decreasing.

b. Find the local minimum value of f

c. Find the inflection points (2)

d.Find the interval on which f is concave up and down

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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a)  You need to perform the first derivative test, hence you need to find f'(x) such that:

`f'(x) = (4x^3)/(x^4+27)`

You need to solve f'(x) = 0 such that:

`(4x^3)/(x^4+27) = 0 =gt 4x^3 = 0 =gt x = 0`

You need to select a value smaller than 0 such that:

`f'(-1) = -4/28 lt 0`

You need to select a value larger than 0 such that:

`f'(1) = 4/28 gt 0`

Hence, the function increases over `(0,oo)`  and it decreases over `(-oo,0).`

b) The function reaches is minimum at x = 0, hence `f(0) = ln 27` .

c) You need to solve f''(x)=0 to find inflection point such that:

`f''(x) = (12x^2(x^4+27) - 4x^3*4x^3)/((x^4+27)^2)`

`f''(x) = (12x^6 + 324x^2 - 16x^6)/((x^4+27)^2)`

`f''(x) = (324x^2 - 4x^6)/((x^4+27)^2)`

`324x^2 - 4x^6 = 0 =gt x^2(324 - 4x^4) = 0`

`x = 0`

`(18-2x^2)(18+2x^2) = 0`

`18-2x^2 = 0 =gt x^2 = 9 =gt x_(3,4) = +-3`

The function has inflection points at x = -3, x = 0 and x = 3.

d) Hence, the function is concave down if `x in (-oo,-3)U(3,oo)`  and it is concave up if `x in (-3,0)U(0,3).`

 

Here is a video explaining the first and second derivative tests:

 

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