A value of c for which conclusion of Mean Value Theorem holds for the function f(x) = log_e x on the

Q: Why do we need the interval to be open $(1, 3)$ for the derivative?

The Mean Value Theorem requires the function to be differentiable on the open interval $(a, b)$ because the derivative at the endpoints might not exist or might not be relevant to the existence of the tangent line within the interval. The theorem guarantees the existence of a point *inside* the interval where the slope matches.

Q: Can I use the Mean Value Theorem if the function isn't differentiable at one point?

No, the Mean Value Theorem strictly requires continuity on the closed interval $[a, b]$ and differentiability on the open interval $(a, b)$. If the function is not differentiable at even one point in $(a, b)$, the theorem's conclusion is not guaranteed.

Q: How do I handle the log properties when solving for $c$?

You have $\frac{1}{c} = \frac{\log_e 3}{2}$. Taking the reciprocal gives $c = \frac{2}{\log_e 3}$. Using the change of base formula $\frac{1}{\log_a b} = \log_b a$, we can rewrite $\frac{1}{\log_e 3}$ as $\log_3 e$, leading to $c = 2 \log_3 e$.

Question

A value of $c$ for which conclusion of Mean Value Theorem holds for the function $f(x) = \log_e x$ on the interval $[1, 3]$ is

Options:

(A) $\log_3 e$

(B) $\log_e 3$

(C) 2 \log_3 e

(D) $\frac{1}{2} \log_3 e$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a vast, smooth, and gently rising landscape, represented by the curve $f(x) = \log_e x$. You are looking at the interval from $x = 1$ to $x = 3$.

We start by identifying our boundary points. At $x = 1$, the function value is $f(1) = \log_e 1 = 0$.

At $x = 3$, the function value is $f(3) = \log_e 3$. These two points, $A(1, 0)$ and $B(3, \log_e 3)$, define the landscape we are traversing.

### The Secant Line: The Average Perspective

Now, imagine drawing a straight, dashed green line connecting point $A$ and point $B$. This is the secant line, which represents the average rate of change of our function over the interval $[1, 3]$.

The slope of this line is calculated as the change in $y$ divided by the change in $x$:

$$Slope = \frac{f(3) - f(1)}{3 - 1}$$

Substituting our values, we get:

$$Slope = \frac{\log_e 3 - 0}{3 - 1} = \frac{\log_e 3}{2}$$

This value is the average slope, the steady, unwavering climb from $A$ to $B$.

### The Tangent Line: The Instantaneous Perspective

The curve is not a straight line; it is a living, breathing logarithmic function. Lagrange's Mean Value Theorem (LMVT) tells us there must be at least one point $c$ in the open interval $(1, 3)$ where the instantaneous slope of the curve is exactly equal to the average slope.

To find this, we need the derivative of our function. The derivative of $f(x) = \log_e x$ is a fundamental result in calculus:

$$f'(x) = \frac{1}{x}$$

At our unknown point $c$, the slope of the tangent is $f'(c) = \frac{1}{c}$.

### The Bridge: Equating the Two

We are now at the heart of the problem. We equate the instantaneous slope to the average slope:

$$f'(c) = \frac{f(3) - f(1)}{3 - 1}$$

This gives us the equation:

$$\frac{1}{c} = \frac{\log_e 3}{2}$$

This is the bridge between the average and the instantaneous. It is a simple algebraic equation that holds the secret to the curve's behavior.

### The Logarithmic Twist: Solving for $c$

Solving for $c$ is where we apply our logarithmic toolkit. Rearranging the equation, we get:

$$c = \frac{2}{\log_e 3}$$

Using the change of base property, where $\frac{1}{\log_a b} = \log_b a$, we can transform this into:

$$c = 2 \log_3 e$$

This is our **final answer**. If you calculate its decimal value, $c \approx 1.82$, which sits perfectly within our interval $(1, 3)$.

We have successfully found the point where the tangent is perfectly parallel to the secant line, confirming the power and elegance of the Mean Value Theorem.

Animated Solution for Mathematics - Differentiation: A value of $c$ for which conclusion of Mean Value Theorem holds for the function $f (x) = lo g_{e} x$ on the interval $[1, 3]$ is

Select Answer:

Visualized Solution

Visualizing the Function and Interval

Lagrange's Mean Value Theorem (LMVT)

Calculating the Secant Slope

Evaluating $f (1)$ and $f (3)$

Differentiating $f (x) = lo g_{e} x$

Equating $f^{'} (c)$ to the Secant Slope

Solving for the Value of $c$

Verification and Geometric Meaning

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Secant Line

The Tangent Line

The Bridge

The Logarithmic Twist

Similar Questions

The value of $c$ in Lagrange’s mean value theorem for the function $f (x) = x^{3} - 4 x^{2} + 8 x + 11$ , where $x \in [0, 1]$ is :

The value of $c$ in the Lagrange's mean value theorem for the function $f (x) = x^{3} - 4 x^{2} + 8 x + 11$ when $x \in [0, 1]$ is :

If $c$ is a point at which Rolle's theorem holds for the function, $f (x) = lo g_{e} (\frac{x ^{2} + α}{7 x})$ in the interval $[3, 4]$ , where $α \in R$ , then $f^{''} (c)$ is equal to :

If $f (x)$ is twice differentiable and continuous function in $x \in [a, b]$ also $f^{'} (x) > 0$ and $f^{''} (x) < 0$ and $c \in (a, b)$ then $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than

If $f (x)$ and $g (x)$ are differentiable function for $0 \leq x \leq 1$ such that $f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2$ , then show that there exist $c$ satisfying $0 < c < 1$ and $f^{'} (c) = 2 g^{'} (c)$ .

If $f (x) = x^{α} lo g x$ and $f (0) = 0$ , then the value of $α$ for which Rolle's theorem can be applied in $[0, 1]$ is

Let $f$ be any function continuous on $[a, b]$ and twice differentiable on $(a, b)$ . If for all $x \in (a, b)$ , $f^{'} (x) > 0$ and $f^{''} (x) < 0$ , then for any $c \in (a, b)$ , $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than:

Let $S$ be the set of all function $f : [0, 1] \to R$ , which are continuous on $[0, 1]$ and differentiable on $(0, 1)$ . Then for every $f$ in $S$ , there exists a $c \in (0, 1)$ depending on $f$ , such that :

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f (0) = 2 = g (1), g (0) = 0$ and $f (1) = 6$ , then for some $c \in] 0, 1 [$

Let $S$ be the set of all functions $f : [0, 1] \to R$ , which are continuous on $[0, 1]$ and differentiable on $(0, 1)$ . Then for every $f$ in $S$ , there exists a $c \in (0, 1)$ , depending on $f$ , such that:

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A value of c for which conclusion of Mean Value Theorem holds for the function f(x)=loge​x on the interval [1,3] is

Select Answer:

Visualized Solution

Visualizing the Function and Interval

Lagrange's Mean Value Theorem (LMVT)

Calculating the Secant Slope

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(3)

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=loge​x

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(c) to the Secant Slope

Solving for the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c