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 use $f'(x) = 2$ instead of $f'(x) \leq 2$?

Great question! The problem gives us an inequality, not an equality. The derivative is *at most* 2, meaning the function can be flatter (have a smaller slope) but never steeper. By using the maximum possible slope, we are finding the *upper bound* of the function's value. If the slope were smaller, the function would be lower, so the upper bound remains valid.

Q: Why did we choose the intervals $[-7, 0]$ and $[-7, -1]$?

We need to find the range for $f(-1) + f(0)$. We are given the value at $x = -7$. To use LMVT effectively, we need to connect our known point ($-7$) to the points we are interested in ($-1$ and $0$). Thus, the intervals $[-7, -1]$ and $[-7, 0]$ are the most direct paths to finding the bounds for $f(-1)$ and $f(0)$ respectively.

Q: Is the lower bound really $-\infty$?

Yes, it is! Since the derivative is only bounded from above ($f'(x) \leq 2$), there is no restriction on how small the derivative can be. The function could have a very large negative slope, allowing it to drop to arbitrarily small values. Therefore, the lower bound is indeed negative infinity.

Question

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

Options:

(A) $[-6, 20]$

(B) $(-\infty, 20]$

(C) $(-\infty, 11]$

(D) $[-3, 11]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are driving a car on a long, straight road. You know your starting position at time $t = -7$, and you know that your speedometer never exceeds $2$ km/h.

If someone asks, "What is the furthest point you could possibly reach by time $t = 0$?", you would intuitively calculate your maximum possible distance based on that speed limit.

This is exactly what we are doing with the function $f(x)$ in this problem. We are given that $f(-7) = -3$ and that the slope, or the derivative $f'(x)$, is always less than or equal to $2$. This is our speed limit.

### The Power of Lagrange's Mean Value Theorem

To turn this intuition into rigorous mathematics, we use the **Lagrange's Mean Value Theorem (LMVT)**. This theorem is a bridge between the average rate of change and the instantaneous rate of change.

It states that for a continuous and differentiable function on an interval $[a, b]$, there exists some point $c$ in that interval such that the slope of the secant line connecting the endpoints is equal to the derivative at $c$:

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

Since we know $f'(c) \leq 2$, we can immediately say that:

$$
\frac{f(b) - f(a)}{b - a} \leq 2
$$

This inequality serves as our master key for the problem.

### Unlocking the Bounds

First, let's find the maximum possible value for $f(0)$. We apply LMVT on the interval $[-7, 0]$. Our starting point is $a = -7$ and our target is $b = 0$.

The formula gives us:

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

Substituting $f(-7) = -3$, we get:

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

With a little algebraic manipulation, this becomes $f(0) + 3 \leq 14$, which simplifies to **$f(0) \leq 11$**. The function cannot climb higher than $11$ at $x = 0$.

Next, we need the bound for $f(-1)$. We repeat the process, this time using the interval $[-7, -1]$. The setup is:

$$
\frac{f(-1) - f(-7)}{-1 - (-7)} \leq 2
$$

Substituting $f(-7) = -3$ again, we get:

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

Multiplying by $6$ gives $f(-1) + 3 \leq 12$, which simplifies to **$f(-1) \leq 9$**.

### The Final Synthesis

Now, we have two solid pieces of information: $f(0) \leq 11$ and $f(-1) \leq 9$. The question asks for the range of the sum $f(-1) + f(0)$.

Since both inequalities are upper bounds, we can simply add them together:

$$
f(-1) + f(0) \leq 9 + 11
$$

This results in **$f(-1) + f(0) \leq 20$**.

Because there is no lower bound on the derivative, the function can decrease without limit, meaning the sum can be any value less than or equal to $20$. In interval notation, this is **$(-\infty, 20]$**.

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 all $x \in (- 7, 0)$ , then for all such functions $f$ , $f (- 1) + f (0)$ lies in the interval:

Select Answer:

Visualized Solution

Visualizing the Constraints

The Tool: Lagrange's Mean Value Theorem

Applying LMVT on $[- 7, 0]$

Finding the Bound for $f (0)$

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

Finding the Bound for $f (- 1)$

Summing the Results

Final Interval and Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Power of Lagrange's Mean Value Theorem

Unlocking the Bounds

The Final Synthesis

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Select Answer:

Visualized Solution

Visualizing 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; } [−7,0]

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