4.5-4.9) password: Determine whether Rolles Theorem applies to the following function on the given interval. If $$ g(x)=x^{3}+x^{2}-8 x-12 i[-2,3] $$

Question

UpStudy AI · Accepted Answer

To determine whether **Rolle's Theorem** applies to the function $$ g(x) = x^3 + x^2 - 8x - 12 $$ on the interval $$[-2, 3]$$, we'll systematically verify the theorem's conditions.

### **Rolle's Theorem Statement**

Rolle's Theorem states that if a function $$ f $$ satisfies the following three conditions on a closed interval $$[a, b]$$:

1. **Continuity:** $$ f $$ is continuous on $$[a, b]$$.
2. **Differentiability:** $$ f $$ is differentiable on $$(a, b)$$.
3. **Equal Endpoints:** $$ f(a) = f(b) $$.

**Then, there exists at least one number $$ c $$ in $$(a, b)$$ such that:**
$$ f'(c) = 0 $$

### **Verifying the Conditions for $$ g(x) $$**

1. **Continuity:**
   - **Polynomials** are continuous everywhere on $$\mathbb{R}$$.
   - **Conclusion:** $$ g(x) $$ is continuous on $$[-2, 3]$$.

2. **Differentiability:**
   - **Polynomials** are differentiable everywhere on $$\mathbb{R}$$.
   - **Conclusion:** $$ g(x) $$ is differentiable on $$(-2, 3)$$.

3. **Equal Endpoints:**
   - Compute $$ g(-2) $$ and $$ g(3) $$:
     $$
     \begin{align*}
     g(-2) &= (-2)^3 + (-2)^2 - 8(-2) - 12 \
           &= -8 + 4 + 16 - 12 \
           &= 0 \
     g(3) &= (3)^3 + (3)^2 - 8(3) - 12 \
           &= 27 + 9 - 24 - 12 \
           &= 0 \
     \end{align*}
     $$
   - Since $$ g(-2) = g(3) = 0 $$, the third condition is satisfied.

### **Conclusion**

All three conditions of Rolle's Theorem are satisfied for the function $$ g(x) = x^3 + x^2 - 8x - 12 $$ on the interval $$[-2, 3]$$. Therefore, **Rolle's Theorem does apply** to this function on the given interval.

### **Additional Insight (Optional): Finding the Point $$ c $$**

While the question doesn't explicitly ask for it, finding the value of $$ c $$ where $$ g'(c) = 0 $$ can provide deeper understanding.

1. **Find the derivative $$ g'(x) $$:**
   $$
   g'(x) = 3x^2 + 2x - 8
   $$

2. **Set the derivative equal to zero and solve for $$ x $$:**
   $$
   3x^2 + 2x - 8 = 0
   $$
   - **Using the quadratic formula:**
     $$
     x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 3 \cdot (-8)}}{2 \cdot 3} = \frac{-2 \pm \sqrt{4 + 96}}{6} = \frac{-2 \pm \sqrt{100}}{6} = \frac{-2 \pm 10}{6}
     $$
   - **Solutions:**
     $$
     x = \frac{8}{6} = \frac{4}{3} \quad \text{and} \quad x = \frac{-12}{6} = -2
     $$
   - **Valid $$ c $$ in $$(-2, 3)$$:**
     - $$ x = \frac{4}{3} $$ lies within the interval $$(-2, 3)$$.
     - $$ x = -2 $$ is an endpoint and not within $$(-2, 3)$$.

3. **Conclusion:**
   - The point $$ c = \frac{4}{3} $$ is where $$ g'(c) = 0 $$ within the interval $$(-2, 3)$$.

### **Final Statement**

**Yes**, Rolle's Theorem applies to $$ g(x) = x^3 + x^2 - 8x - 12 $$ on the interval $$[-2, 3]$$ because:

1. $$ g(x) $$ is continuous on $$[-2, 3]$$.
2. $$ g(x) $$ is differentiable on $$(-2, 3)$$.
3. $$ g(-2) = g(3) $$.

As a result, there exists at least one $$ c $$ in $$(-2, 3)$$ (specifically, $$ c = \frac{4}{3} $$) such that $$ g'(c) = 0 $$.
Rolle's Theorem applies to the function $$ g(x) = x^3 + x^2 - 8x - 12 $$ on the interval $$[-2, 3]$$ because the function is continuous and differentiable on this interval, and $$ g(-2) = g(3) $$. Therefore, there is at least one number $$ c $$ in $$(-2, 3)$$ where the derivative $$ g'(c) = 0 $$.

4.5-4.9) password: Determine whether Rolles Theorem applies to the following function on the given interval. If \( g(x)=x^{3}+x^{2}-8 x-12 i[-2,3] \)

Solución de tutoría real

Responder

Solución

The Deep Dive

preguntas relacionadas

Latest Calculus Questions