For the function f(x) = x^4 - 5x^3find intervals where f(x) is increasing or decreasing, points of local maxima and minima of f(x), intervals where f(x) is concave up or concave down. the inflection points of f(x).Use a comma to separate points. Enter 'NONE if there are none.a. Interval (s) where f(x) is increasing: b. Interval (s) where f(x) is decreasing:c. Point(s) of local maxima of f(x):d. Point(s) of local minima of f(x)

Respuesta :

Given:

[tex]f(x)=x^4-5x^3[/tex]

Aim:

We need to find intervals where f(x) is increasing or decreasing, points of local maxima and minima of f(x), and intervals where f(x) is concave up or concave down. the inflection points of f(x).

Explanation:

Differentiate the given function with respect to x to find the interval of increasing and decreasing.

[tex]f^{\prime}(x)=4x^3-5\times3x^2[/tex]

[tex]f^{\prime}(x)=4x^3-15x^2[/tex][tex]Let\text{ }\times f^{\prime}(x)=0\text{ to find a critical point.}[/tex]

[tex]4x^3-15x^2=0[/tex]

[tex]x^2(4x-15)=0[/tex][tex]x=0\text{ and 4x-15=0}[/tex]

[tex]x=0\text{ and x=}\frac{15}{4}.[/tex]

We find the following values for the variation chart.

[tex]-\infty<-1<0[/tex]

[tex]f^{\prime}(-1)=4(-1)^3-15(-1)^2=-19=negative[/tex]

[tex]-1<1<\frac{15}{4}[/tex][tex]f^{\prime}(1)=4(1)^3-15(1)^2=-11=negative[/tex]

[tex]\frac{15}{4}<4<\infty[/tex][tex]f^{\prime}(4)=4(4)^3-15(4)^2=16=positive[/tex]

Consider the variation chart.

a)

[tex]\text{ The interval of increasing is x}\in(\frac{15}{4},\infty).[/tex]

b)

[tex]\text{ The interval of decreasing is x}\in(-\infty,\frac{15}{4})[/tex]

Ver imagen HoltenW305626