Determine whether or not the function $f ( x ) = e ^ { x } \left( x ^ { 2 } - 3 x \right)$ satisfies the hypothesis of Rolle's Theorem on the interval [0, 3]. If it does, find the exact values of all $c \in ( 0,3 )$ that satisfy the conclusion of Rolle's Theorem.

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Rolle's Theorem states that if a function $f$ is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c$ in $(a, b)$ such that $f'(c) = 0$. For the function $f(x) = e^x(x^2 - 3x)$ on the interval $[0, 3]$:1. **Continuity and Differentiability**: The function $f(x) = e^x(x^2 - 3x)$ is continuous and differentiable on $[0, 3]$ and indeed on all of $\mathbb{R}$, because it is the product of two functions ($e^x$ and $x^2 - 3x$) that are both continuous and differentiable everywhere.2. **Equal Endpoints**: $f(0) = e^0(0^2 - 3\cdot0) = 0$ and $f(3) = e^3(3^2 - 3\cdot3) = e^3(9 - 9) = 0$. Thus, $f(0) = f(3)$.Since the function satisfies all the conditions of Rolle's Theorem, we can conclude that there exists at least one $c$ in $(0, 3)$ such that $f'(c) = 0$.To find $c$, we differentiate $f(x)$ to get $f'(x) = e^x(x^2 - 3x) + e^x(2x - 3) = e^x(x^2 - x - 3)$. Setting $f'(x) = 0$ gives $e^x(x^2 - x - 3) = 0$. Since $e^x$ is never 0, we solve $x^2 - x - 3 = 0$. The solutions to this quadratic equation are $x = \frac{1 \pm \sqrt{1 + 4\cdot3}}{2} = \frac{1 \pm \sqrt{13}}{2}$. Only $x = \frac{1 + \sqrt{13}}{2}$ lies in the interval $(0, 3)$, so this is the value of $c$ that satisfies the conclusion of Rolle's Theorem.

Determine Whether or Not the Function f(x)=ex(x2−3x)f ( x ) = e ^ { x } \left( x ^ { 2 } - 3 x \right)f(x)=ex(x2−3x)

Determine Whether or Not the Function $f ( x ) = e ^ { x } \left( x ^ { 2 } - 3 x \right)$