Chapter 3: Q. 91 (page 277)

Prove part (b) of Theorem 3.10: If both f and $f'$ are differentiable on an interval I, and $f''$ is negative on I, then f is concave down on I.

Short Answer

Expert verified

Since $f''$ is negative $f'$ is decreasing on I. Therefore, from the definition of concavity f is concave down.

Step by step solution

Step 1. Given Information.

Given: $f$ and $f'$ are differentiable on I and $f''$ is negative.

Step 2. Theorem

The Derivative Measures Where a Function is Increasing or Decreasing

Let f be a function that is differentiable on an interval I.

(a) If $f'$ is positive in the interior of I, then f is increasing on I.

(b) If $f'$ is negative in the interior of I, then f is decreasing on I.

Step 3. Proof.

Now we know $f$ and $f'$ are differentiable on I and $f''$ is negative on I.

So, from the above theorem it can be concluded that $f'$ is decreasing on I since $f''$ is negative.

Therefore, from the definition of concavity $f$ is concave down since $f'$ is decreasing.

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Prove part (b) of Theorem 3.10: If both f and $f'$ are differentiable on an interval I, and $f''$ is negative on I, then f is concave down on I.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Theorem

Step 3. Proof.

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Prove part (b) of Theorem 3.10: If both f and f'are differentiable on an interval I, and f''is negative on I, then f is concave down on I.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Theorem

Step 3. Proof.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Discrete Mathematics

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Theoretical and Mathematical Physics

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Calculus

Study anywhere. Anytime. Across all devices.

Prove part (b) of Theorem 3.10: If both f and $f'$ are differentiable on an interval I, and $f''$ is negative on I, then f is concave down on I.