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Solution:
Given function: \( f(x) = \sqrt{x} = x^{1/2} \)
Differentiating with respect to \( x \):
$$ f'(x) = \frac{d}{dx}\left(x^{1/2}\right) = \frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2\sqrt{x}} \quad \text{--- (1)} $$
Applying Lagrange's Mean Value Theorem for interval \( [4, 9] \):
$$ f(b) - f(a) = (b - a)f'(c) \quad \text{where } c \in (4, 9) $$
$$ f(9) - f(4) = (9 - 4)f'(c) $$
$$ \sqrt{9} - \sqrt{4} = 5 \cdot \frac{1}{2\sqrt{c}} $$
$$ 3 - 2 = \frac{5}{2\sqrt{c}} \implies 1 = \frac{5}{2\sqrt{c}} $$
$$ 2\sqrt{c} = 5 \implies 4c = 25 $$
Solution:
Given function: \( f(x) = Ax^2 + Bx + C \)
Differentiating with respect to \( x \):
$$ f'(x) = \frac{d}{dx}(Ax^2 + Bx + C) = 2Ax + B \quad \text{--- (1)} $$
Applying Lagrange's Mean Value Theorem in \( [a, b] \):
$$ f(b) - f(a) = (b - a)f'(c) $$
$$ (Ab^2 + Bb + C) - (Aa^2 + Ba + C) = (b - a)(2Ac + B) $$
$$ A(b^2 - a^2) + B(b - a) = (b - a)(2Ac + B) $$
Factoring out \( (b - a) \) from the left-hand side:
$$ (b - a)[A(b + a) + B] = (b - a)(2Ac + B) $$
Since \( a \neq b \implies (b - a) \neq 0 \), dividing both sides by \( (b - a) \):
$$ 2Ac + B = A(a + b) + B $$
$$ 2Ac = A(a + b) \implies 2c = a + b $$
Solution:
Lagrange's Mean Value Theorem states that if \( f(x) \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \), then there exists at least one \( c \in (a, b) \) such that:
$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$
Differentiating \( f(x) = 4(6 - x)^{2/3} \):
$$ f'(x) = 4 \cdot \frac{2}{3}(6 - x)^{\frac{2}{3} - 1} \cdot \frac{d}{dx}(6 - x) $$
$$ f'(x) = \frac{8}{3}(6 - x)^{-1/3} \cdot (-1) = -\frac{8}{3(6 - x)^{1/3}} $$
Notice that at \( x = 6 \in (5, 7) \), the denominator becomes zero, causing \( f'(x) \) to become undefined.
Solution:
By Lagrange's Mean Value Theorem, for any two points in \( [a, b] \), there exists some \( c \in (a, b) \) such that:
$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$
Since it is given that \( f'(x) < 0 \) everywhere in the interval:
$$ f'(c) < 0 \implies \frac{f(b) - f(a)}{b - a} < 0 $$
Because \( a < b \), we have \( (b - a) > 0 \). Multiplying both sides by the positive quantity \( (b - a) \):
$$ f(b) - f(a) < 0 \implies f(b) < f(a) $$
Solution:
The trigonometric function \( f(x) = \sin x \) is continuous in \( [x_1, x_2] \) and differentiable in \( (x_1, x_2) \).
Differentiating \( f(x) \):
$$ f'(x) = \frac{d}{dx}(\sin x) = \cos x $$
According to Lagrange's Mean Value Theorem, there exists at least one \( c \in (x_1, x_2) \) such that:
$$ \cos c = \frac{f(x_2) - f(x_1)}{x_2 - x_1} = \frac{\sin x_2 - \sin x_1}{x_2 - x_1} $$
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