Question

If \[ f(x)=x^3-19x+30 \] is a real valued function with domain \([-4,1]\), then the value of \(c\) according to Lagrange's Mean Value Theorem for \(f(x)\) is

• A. \(\sqrt{4.33}\)
• B. \(-2\)
• C. \(-\sqrt{4.33}\)
• D. \(-3\)

Hint:

For LMVT problems, first compute the average rate of change \[ \frac{f(b)-f(a)}{b-a}, \] then equate it to \(f'(c)\). Finally, choose the solution that lies inside the open interval \((a,b)\).

Answer 1

The Correct Option is C

Concept: Lagrange's Mean Value Theorem states that if a function is:
• Continuous on \([a,b]\),
• Differentiable on \((a,b)\), then there exists a point \(c\in(a,b)\) such that \[ f'(c) = \frac{f(b)-f(a)}{b-a}. \] Since a polynomial is continuous and differentiable everywhere, LMVT can be applied directly.

Step 1: Compute \(f(1)\). \[ f(1) = 1-19+30 = 12. \]

Step 2: Compute \(f(-4)\). \[ f(-4) = (-4)^3-19(-4)+30. \] \[ = -64+76+30. \] \[ = 42. \]

Step 3: Find the average rate of change. \[ \frac{f(1)-f(-4)} {1-(-4)} = \frac{12-42}{5}. \] \[ = \frac{-30}{5} = -6. \] Thus LMVT gives \[ f'(c)=-6. \]

Step 4: Differentiate \(f(x)\). \[ f'(x) = 3x^2-19. \] Therefore, \[ 3c^2-19=-6. \] \[ 3c^2=13. \] \[ c^2=\frac{13}{3}. \] Hence, \[ c=\pm\sqrt{\frac{13}{3}}. \]

Step 5: Select the value lying in \((-4,1)\). Numerically, \[ \sqrt{\frac{13}{3}} \approx \sqrt{4.33}. \] The two values are \[ c=\sqrt{4.33} \quad\text{and}\quad c=-\sqrt{4.33}. \] Since LMVT requires \[ c\in(-4,1), \] only \[ -\sqrt{4.33} \] belongs to the interval. Therefore, \[ \boxed{-\sqrt{4.33}}. \]