Question

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

Short answer

Since $f\text{'}\text{'}$is negative $f\text{'}$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 $f\text{'}$are differentiable on I and $f\text{'}\text{'}$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 $f\text{'}$is positive in the interior of I, then f is increasing on I.

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

(c) If $f\text{'}$is zero in the interior of I, then f is constant on I.

Step 3. Proof.

Now we know $f$and $f\text{'}$are differentiable on I and $f\text{'}\text{'}$is negative on I.

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

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