Question

Prove Theorem 3.11 (b): If $x=c$is a critical point of $f$, both $f$and $f\text{'}$are differentiable near $x=c$, and if $f\text{'}\text{'}$(c) is negative, then $f$has a local maximum at $x=c$.

Short answer

If $f\text{'}\left(c\right)=0$and $f\text{'}\text{'}\left(c\right)<0$then $f$has a local maximum at $x=c$.

Because when second derivative is negative the first derivative will be decreasing near $x=c$.

And the first derivative is the slope of the function and slope will only decrease if the point is the local maxima.

Step-by-step solution

Step 1. Given Information.

$x=c$is a critical point of $f$which means:

$f\text{'}\left(c\right)=0$and given $f\text{'}\text{'}\left(c\right)<0$.

And, $f\text{'}andf\text{'}\text{'}$both are differentiable near c which means both are continuous.

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 from above theorem suppose we are applying it for $f\text{'}$. Then if $f\text{'}\text{'}$is negative which is given then $f\text{'}$is decreasing in the interval. This means the slope of the function is decreasing and if the slope is decreasing this means the point at which the slope is 0 will be the point of local maxima as before that point the slope will be positive and just after the point it will become negative.

That's why the point $x=c$will be the local maxima.