Question

The derivative of f(x)=(x^4/3)-(x^5/5) attains its maximum at x= ? ...?

Answer 1

The value of x is $\boxed{\frac{4}{3}}$ for which the derivative of  $f\left( x \right) =\dfrac{{{x^4}}}{3} - \dfrac{{{x^5}}}{5}$ attains the maximum.

Further explanation:

Given:

The function is $f\left( x \right) =\dfrac{{{x^4}}}{3} - \dfrac{{{x^5}}}{5}.$

Explanation:

The given function is $f\left( x \right)=\dfrac{{{x^4}}}{3} - \dfrac{{{x^5}}}{5}.$

Differentiate the above equation with respect to x.

$\begin{aligned}\frac{d}{{dx}}f\left( x \right) &= \frac{d}{{dx}}\left( {\frac{{{x^4}}}{3} - \frac{{{x^5}}}{5}} \right)\\&= \frac{{4{x^3}}}{3} - \frac{{5{x^4}}}{5}\\&= \frac{{4{x^3}}}{3} - {x^4}\\\end{aligned}$

Again differentiate with respect to x.

$\begin{aligned}\frac{{{d^2}}}{{d{x^2}}}f\left( x \right) &= \frac{{{d^2}}}{{d{x^2}}}\left( {\frac{{4{x^3}}}{3} - {x^4}} \right)\\&=\frac{{3 \times 4{x^2}}}{3} - 4{x^3}\\&= 4{x^2} - 4{x^3}\\\end{aligned}$

Substitute the first derivative equal to zero.

$\begin{aligned}\frac{d}{{dx}}f\left( x \right)&= 0\\\frac{{4{x^3}}}{3} - {x^4}&= 0\\\frac{{4{x^3}}}{3} &= {x^4}\\\frac{4}{3}&= \frac{{{x^4}}}{{{x^3}}}\\\frac{4}{3}&= x\\\end{aligned}$

The value of x is $\boxed{\frac{4}{3}}$ for which the derivative of $f\left( x \right)=\dfrac{{{x^4}}}{3} - \dfrac{{{x^5}}}{5}$ attains the maximum.

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Application of derivatives

Keywords: Derivative, attains, maximum, value of x, function, differentiate, minimum value.

Answer 2

In order to get the maximum (or minimum) of a curve, we need to get the first derivative of the function or the slope then equate to zero. In order to confirm if the value of x is really the maxima, we can proceed to the second derivative and substitute the x to the equation. If the result is a negative number, then the point is the maxima. Therefore,

f(x) = (x^4/3) - (x^5/5)
f'(x) = (4x^3/3) - x^4 = 0

thus x = 4/3

check if the point is really the maxima

f"(x) = 4x^2 - 4x^3

substituting x = 4/3, we get an answer of -2.3704 thus the point is a maxima.