show that the function ( f(x)=\frac{1}{3}x^{3}-2x^{2}+9x ) has no relative extreme points. relative extreme…

Explanation:

Step1: Find the first - derivative

Given $f(x)=\frac{1}{3}x^{3}-2x^{2}+9x$. Using the power - rule $\frac{d}{dx}(x^{n}) = nx^{n - 1}$, we have $f^\prime(x)=x^{2}-4x + 9$.

Step2: Set the first - derivative equal to zero

We want to find the critical points, so we solve the quadratic equation $x^{2}-4x + 9=0$. The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 1$, $b=-4$, and $c = 9$.

Step3: Calculate the discriminant

The discriminant $\Delta=b^{2}-4ac=(-4)^{2}-4\times1\times9=16 - 36=-20$.

Step4: Analyze the result

Since $\Delta=-20<0$, the quadratic equation $x^{2}-4x + 9 = 0$ has no real solutions. Relative extreme points occur when $f^\prime(x)=0$. Since there are no real values of $x$ for which $f^\prime(x)=0$, the function $f(x)=\frac{1}{3}x^{3}-2x^{2}+9x$ has no relative extreme points.

Answer:

Relative extreme points exist when $f^\prime(x)=0$. In this case, because the discriminant of $x^{2}-4x + 9 = 0$ is negative ($\Delta=-20<0$), the function $f(x)=\frac{1}{3}x^{3}-2x^{2}+9x$ has no relative extreme points.