Let f be differentiable for all x. If f(1) = -2 and f'(x) = 2 for x in [1, 6], then (A) f(6) = 8 (

Q: Wait, why do we use the Mean Value Theorem instead of just integrating $f'(x)$?

That is a fantastic question! You absolutely could use the Fundamental Theorem of Calculus: $f(6) = f(1) + \int_1^6 f'(x) dx$. Since $f'(x) \geq 2$, the integral $\int_1^6 f'(x) dx$ must be $\geq \int_1^6 2 dx = 10$. Both methods lead to the same result! The MVT is just a more direct, algebraic way to express the same physical reality.

Q: Is it possible for $f(6)$ to be exactly 8?

Yes, it is! If $f'(x) = 2$ for all $x \in [1, 6]$, then $f(x)$ would be a straight line with a slope of 2, specifically $f(x) = 2x - 4$. At $x=1$, $f(1) = 2(1) - 4 = -2$, and at $x=6$, $f(6) = 2(6) - 4 = 8$. So, 8 is the absolute minimum boundary.

Q: Does the function have to be differentiable for this to work?

Yes, the differentiability is crucial. The Mean Value Theorem requires the function to be continuous on the closed interval and differentiable on the open interval. Without differentiability, the slope could jump around, and the MVT wouldn't be guaranteed to hold. Always check your conditions!

Question

Let $f$ be differentiable for all $x$. If $f(1) = -2$ and $f'(x) \geq 2$ for $x \in [1, 6]$, then

Options:

(A) f(6) \geq 8

(B) f(6) < 8

(C) f(6) < 5

(D) f(6) = 5

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear students! Today, we are going to peel back the layers of a beautiful calculus problem. It is not just about solving for $f(6)$; it is about understanding the soul of a function.

Imagine you are standing at the starting line of a race. You are at $x=1$, and your position is $f(1) = -2$. You have a coach who tells you, "No matter what happens, your speed—your slope—must be at least $2$."

How high can you climb by the time you reach $x=6$?

### Visualizing the Climb

First, let us ground ourselves. We are looking at the interval $x \in [1, 6]$. We know our starting point is $(1, -2)$. This is our anchor.

The constraint $f'(x) \geq 2$ is the most exciting part. In the world of calculus, $f'(x)$ is the instantaneous rate of change. If $f'(x) \geq 2$, it means our function is forced to climb.

It cannot rest, it cannot dip, and it cannot grow slowly. It is a relentless, upward march. I want you to visualize a dashed line with a slope of $2$ starting from our point. That line represents the absolute minimum path our function must take. Anything below that line is forbidden territory.

### The Bridge: Lagrange's Mean Value Theorem

How do we turn this intuition into a rigorous proof? We need a bridge. That bridge is Lagrange's Mean Value Theorem (LMVT).

The theorem tells us that for a differentiable function, there exists some point $c$ in our interval $(1, 6)$ where the instantaneous slope $f'(c)$ is exactly equal to the average slope over the entire interval. Mathematically, this is expressed as:

$$\frac{f(6) - f(1)}{6 - 1} = f'(c)$$

This equation is the heartbeat of the problem. It connects the start and the end of our journey to the behavior of the function in between.

### The Logical Leap

Now, we apply our constraint. We know that $f'(x) \geq 2$ for every single point in the interval. Therefore, at our mysterious point $c$, it must also be true that $f'(c) \geq 2$.

By substituting this into our LMVT equation, we transform a calculus problem into a simple algebraic inequality:

$$\frac{f(6) - f(1)}{6 - 1} \geq 2$$

This is where the magic happens. We are no longer dealing with abstract derivatives; we are dealing with the relationship between the endpoints. Let us plug in our known value, $f(1) = -2$. Be careful with the signs here—it is the most common place for students to stumble!

$$\frac{f(6) - (-2)}{5} \geq 2$$

### The Final Stretch

Simplifying this is a joy. The numerator becomes $f(6) + 2$. Multiplying both sides by $5$, we get:

$$f(6) + 2 \geq 10$$

Subtracting $2$ from both sides, we arrive at our destination:

**$$f(6) \geq 8$$**

There it is! The minimum possible value for our function at $x=6$ is **$8$**.

Geometrically, this means that while our function could end up at $9, 10, 100,$ or higher, it is physically impossible for it to be below $8$. You have successfully used the power of calculus to predict the future of a function based on its rate of growth.

Animated Solution for Mathematics - Differentiation: Let $f$ be differentiable for all $x$ . If $f (1) = - 2$ and $f^{'} (x) \geq 2$ for $x \in [1, 6]$ , then

Select Answer:

Visualized Solution

Visualizing $f (1) = - 2$

The Interval $x \in [1, 6]$

The Constraint $f^{'} (x) \geq 2$

Lagrange's Mean Value Theorem

The LMVT Equation

Applying $f^{'} (c) \geq 2$

Forming the Inequality

Substituting $f (1) = - 2$

Simplifying the Denominator

Multiplying by $5$

Isolating $f (6)$

Conclusion: $f (6) \geq 8$

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

Visualizing the Climb

The Bridge

The Logical Leap

The Final Stretch

Similar Questions

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be differentiable for all x. If f(1)=−2 and f′(x)≥2 for x∈[1,6], then

Select Answer:

Visualized Solution

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)=−2

The Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[1,6]

The Constraint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)≥2

Lagrange's Mean Value Theorem

The LMVT Equation

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(c)≥2

Forming the Inequality

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)=−2