Let the function, f: [-7, 0] → R be continuous on [-7, 0] and differentiable on (-7, 0). If f(-7) =

Q: Wait, why can't we just assume $f(x) = 2x + c$ and solve it like a linear function?

That is a common trap! The condition $f'(x) \le 2$ only gives us an upper bound on the slope. It does not mean the function *must* be a line with slope 2. The function could be much 'flatter' or even decrease rapidly. By using the Mean Value Theorem, we are finding the absolute 'ceiling' the function cannot cross, which is a much more powerful and general approach than assuming a specific shape.

Q: Why is the interval $(-\infty, 20]$ and not something like $[k, 20]$?

Great observation! The constraint $f'(x) \le 2$ only restricts how fast the function can *increase*. It places no restriction on how fast the function can *decrease*. Because the derivative can be any value less than or equal to 2 (including very large negative numbers), the function can drop as low as it wants. Therefore, there is no lower bound, leading us to negative infinity.

Q: Can we use the Fundamental Theorem of Calculus to integrate $f'(x) \le 2$?

You could try, but it's tricky! Integrating $f'(x) \le 2$ from $-7$ to $x$ gives $\int_{-7}^{x} f'(t) dt \le \int_{-7}^{x} 2 dt$, which simplifies to $f(x) - f(-7) \le 2(x + 7)$. This is essentially the same result as the Mean Value Theorem. While integration works, the Mean Value Theorem is the more direct and elegant tool for this specific type of inequality problem.

Question

Let the function, $f: [-7, 0] 	o R$ be continuous on $[-7, 0]$ and differentiable on $(-7, 0)$. If $f(-7) = -3$ and $f' (x) \le 2$, for $x \in (-7, 0)$, then for all such functions $f, f(-1) + f (0)$ lies in the interval:

Options:

(A) [-5, -7]

(B) (-\infty, 6]

(C) (-\infty, 20]

(D) [-5, 3]

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to embark on a journey through a problem that might seem like a simple exercise in inequalities, but is actually a profound lesson in how calculus governs the behavior of functions.

Imagine you are driving a car on a winding mountain road. You know exactly where you started, and you have a strict speed limit. Can you predict the furthest point you could possibly reach? That is the exact physical intuition behind this problem.

We are given a function $f(x)$ defined on the interval $[-7, 0]$. We know it starts at the point $(-7, -3)$. We are also given a 'speed limit' for the function's growth: its derivative, $f'(x)$, is always less than or equal to $2$.

This means the slope of the tangent line at any point is capped. The function simply cannot climb faster than a line with a slope of $2$.

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

To turn this 'speed limit' into a concrete mathematical bound, we need a bridge. That bridge is Lagrange's Mean Value Theorem (LMVT).

The theorem tells us that for any interval $[a, b]$, there exists some point $c$ inside that interval where the instantaneous slope $f'(c)$ is exactly equal to the average slope of the function over that interval. Mathematically, this is expressed as:

$$
\frac{f(b) - f(a)}{b - a} = f'(c)
$$

Geometrically, this means that the average rate of change of our function is constrained by the maximum value of its derivative. If $f'(x) \le 2$, then the average rate of change must also be less than or equal to $2$.

It is like saying if your speedometer never exceeds $100 	ext{ km/h}$, your average speed over any trip cannot possibly exceed $100 	ext{ km/h}$ either.

### Calculating the Bounds

Let us apply this logic to our specific targets: $f(-1)$ and $f(0)$. First, consider the interval $[-7, -1]$. Applying LMVT, we have:

$$
\frac{f(-1) - f(-7)}{-1 - (-7)} = f'(c_1) \le 2
$$

Substituting our known starting value $f(-7) = -3$, the equation becomes:

$$
\frac{f(-1) - (-3)}{6} \le 2
$$

Multiplying both sides by $6$, we get $f(-1) + 3 \le 12$, which simplifies to $f(-1) \le 9$. This is our first milestone! The function cannot be higher than $9$ at $x = -1$.

Now, let us repeat this process for the interval $[-7, 0]$ to find the bound for $f(0)$:

$$
\frac{f(0) - f(-7)}{0 - (-7)} = f'(c_2) \le 2
$$

Again, substituting $f(-7) = -3$:

$$
\frac{f(0) - (-3)}{7} \le 2
$$

This leads to $f(0) + 3 \le 14$, or $f(0) \le 11$. We have successfully established the 'ceiling' for both points.

### The Final Synthesis

The question asks us for the interval of the sum $f(-1) + f(0)$. Since we know $f(-1) \le 9$ and $f(0) \le 11$, it follows logically that their sum must be less than or equal to the sum of their maximums:

$$
f(-1) + f(0) \le 9 + 11 = 20
$$

But what about the lower bound? As we discussed, the derivative is only bounded from above. The function could plummet downwards as sharply as it wants, meaning there is no lower limit to how small the sum can be.

Thus, the sum can be any value from negative infinity up to $20$. Our final interval is **$(-\infty, 20]$**.

This problem is a beautiful reminder that calculus is not just about finding derivatives and integrals; it is about understanding the constraints that shape the world around us. Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic behind the math!

Animated Solution for Mathematics - Differentiation: Let the function, $f : [- 7, 0] \to R$ be continuous on $[- 7, 0]$ and differentiable on $(- 7, 0)$ . If $f (- 7) = - 3$ and $f^{'} (x) \leq 2$ , for $x \in (- 7, 0)$ , then for all such functions $f, f (- 1) + f (0)$ lies in the interval:

Select Answer:

Visualized Solution

Understanding the Constraints

The Tool: Lagrange's Mean Value Theorem

Applying LMVT on $[- 7, - 1]$

Substituting Known Values

Finding the Bound for $f (- 1)$

Applying LMVT on $[- 7, 0]$

Substituting Values for $f (0)$

Finding the Bound for $f (0)$

Summing the Inequalities

The Final Interval

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Tool

Calculating the Bounds

The Final Synthesis

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Select Answer:

Visualized Solution

Understanding the Constraints

The Tool: Lagrange's Mean Value Theorem

Applying LMVT on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [−7,−1]

Substituting Known Values

Finding the Bound for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−1)

Applying LMVT on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [−7,0]

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

Finding the Bound for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)