Question

find the absolute maximum and absolute minimum of the function (f(x)=\frac{x^{3}}{3}-\frac{5x^{2}}{2}+4x + 4) over the interval (0,3). enter an exact answer. if there is more than one value of (x) in the interval at which the maximum or minimum occurs, you should use a comma to separate them. provide your answer below: absolute maximum of (square) at (x = square) absolute minimum of (square) at (x=square)

Explanation:

Step1: Find the derivative of the function

The function is $f(x)=\frac{x^{3}}{3}-\frac{5x^{2}}{2}+4x + 4$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=x^{2}-5x + 4$.

Step2: Set the derivative equal to zero and solve for x

Set $x^{2}-5x + 4 = 0$. Factor the quadratic equation: $(x - 1)(x - 4)=0$. So $x = 1$ or $x = 4$. But $x = 4$ is outside the interval $[0,3]$, so we only consider $x = 1$.

Step3: Evaluate the function at the critical point and endpoints

Evaluate $f(x)$ at $x = 0$, $x = 1$, and $x = 3$.
$f(0)=\frac{0^{3}}{3}-\frac{5\times0^{2}}{2}+4\times0 + 4=4$.
$f(1)=\frac{1^{3}}{3}-\frac{5\times1^{2}}{2}+4\times1 + 4=\frac{1}{3}-\frac{5}{2}+4 + 4=\frac{2 - 15+24 + 24}{6}=\frac{35}{6}\approx5.83$.
$f(3)=\frac{3^{3}}{3}-\frac{5\times3^{2}}{2}+4\times3 + 4=9-\frac{45}{2}+12 + 4=\frac{18-45 + 24 + 8}{2}=\frac{5}{2}=2.5$.

Answer:

Absolute maximum of $\frac{35}{6}$ at $x = 1$; Absolute minimum of $\frac{5}{2}$ at $x = 3$