Mathematics - Application of Derivative Question with Solution | TestHub

The given function is $f(x)=2x^3-15x^2+36x-48$ and $A=\left\{x\mid x^2+20\le 9x\right\}$

Step 1: Define the set A

To define the set

$$A=\{x\mid x^2+20\le 9x\}$$

, we solve the inequality:

$$x^2-9x+20\le 0$$

Factoring the quadratic expression gives:

$$(x-4)(x-5)\le 0$$

The inequality holds for values of $x$ between or equal to the roots 4 and 5. Thus, the set $A$ is the closed interval $[4,5]$.

Step 2: Find critical points

We find the derivative of the function $f(x)=2x^3-15x^2+36x-48$ to locate critical points:

$$f^{\prime }(x)=6x^2-30x+36$$

Set the derivative to zero to find critical points:

$$6x^2-30x+36=0$$

$$x^2-5x+6=0$$

$$(x-2)(x-3)=0$$

The critical points are $x=2$ and $x=3$. Both points lie outside our interval of interest $[4,5]$.

Step 3: Evaluate at endpoints

Since no critical points are within the interval $[4,5]$, the maximum value must occur at one of the endpoints, $x=4$ or $x=5$.

We evaluate $f(x)$ at $x=4$:

$$f(4)=2(4{)}^3-15(4{)}^2+36(4)-48=128-240+144-48=-16$$

We evaluate $f(x)$ at $x=5$:

$$f(5)=2(5{)}^3-15(5{)}^2+36(5)-48=250-375+180-48=7$$

Step 4: Determine maximum value

Comparing the values $f(4)=-16$ and $f(5)=7$, the maximum value is $7$.

Answer: The maximum value of the function $f(x)$ on the set $A$ is $7$.