Lagrange Mean Value Theorem Application
This solution determines the value of x where the Lagrange Mean Value Theorem (LMVT) is satisfied for the function $f(x) = x^3 + 5$ within the interval (1, 4).
Applying the Lagrange Mean Value Theorem
The LMVT states that if a function $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, there exists a $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
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Find the derivative of the function: $f(x) = x^3 + 5$ $f'(x) = 3x^2$
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Calculate the function values at the interval endpoints: $f(1) = 1^3 + 5 = 1 + 5 = 6$ $f(4) = 4^3 + 5 = 64 + 5 = 69$
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Calculate the slope of the secant line between the endpoints: Slope = $\frac{f(4) - f(1)}{4 - 1} = \frac{69 - 6}{3} = \frac{63}{3} = 21$
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Set the derivative equal to the slope and solve for $x$ (denoted as $c$ for the theorem): $f'(c) = 3c^2 = 21$ $c^2 = \frac{21}{3}$ $c^2 = 7$ $c = \pm\sqrt{7}$
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Select the value within the interval (1, 4): The interval is (1, 4). We need $c$ such that $1 < c < 4$. $\sqrt{7} \approx 2.64575$ $-\sqrt{7} \approx -2.64575$ Only $\sqrt{7}$ lies within the interval (1, 4).
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Round the result to two decimal places: $c = \sqrt{7} \approx 2.64575$ Rounded to two decimal places, $c \approx 2.65$. This value falls between 2.6 and 2.7.
Conclusion
The value of $x$ at which the Lagrange Mean Value Theorem is satisfied for $f(x) = x^3 + 5$ in the interval (1, 4) is approximately 2.65.
Important Questions from Mean Value Theorem
A series expansion for the function sin θ is
If f is the derivative of some function on [a, b], then there exists a number c in (a, b) such that Integral of f with respect to x =
Which condition is not required in checking for Taylor's theorem?
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According to the Mean Value Theorem, for a continuous function f(x) in the interval [a, b], there exists a value ξ in this interval such that \(\mathop \smallint \limits_a^b f\left( x \right)dx =\)
