0

Let f(x) be a continuous function such that $f’(x) = (12x – 48)/[3(x – 4)^2 + 1]$ for all real numbers x.

(a) If f(x) attains its minimum value at x = k, find k.

(b) It is given that the extreme value of f(x) = 5. Find f(x) and lim f(x) when x tends to infinity.

From f’(x) = 0, we get x = 4. Using the 1st derivative test, we find the f(x) attains its minimum at x = 4 = k.

To find f(x), we integrate f’(x) and get $f(x) = -2/[3(x – 4)^2 + 1] + C$.

To find C, the answer says “since f(x) has only one extreme value, we have f(4) = 5.”

Then, C = … = 7

When x tends to infinity, lim f(x) = … = 7

My question is (1) how do we know that there is only one extreme value? (2) Why 7 thus found (or the right endpoint) is not an extreme value?

Mick
  • 17,141

1 Answers1

0

Easiest to make change of variable:

$y = (x-4) \implies \frac{dy}{dx} = 1.$

$$f'(y) = \frac{12y}{3y^2 + 1}.$$

It is immediate that at $y = 0, f'(y) = 0$ and $f''(y) > 0.$

This verifies that $f(y)$ has a minimum at $y = 0$.

Further, it is straight forward that

$$f(y) = \int f'(y)dy = \int \frac{12y}{3y^2 + 1} = 2 \log (3y^2 + 1) + C.$$

This means that when $y = 0, 5 = f(y) = C.$

Thus, $f(y) = 2 \log (3y^2 + 1) + 5.$
This immediately allows the conclusion that $f(y)$ has no extreme value
for any finite value of $y$, except at $y = 0$.

Alternatively, you can reason that $f'(y) = 0$
when and only when $y = 0$.

Edit

The following analysis resulted from back and forth comments with Mick, following this answer. I think that this analysis deserves to be included in the answer.

$f(y)$ is continuous throughout $\mathbb{R}.$
$f''(y)$ and $f'(y)$ are both well defined throughout $\mathbb{R}.$
$f''(0) > 0, f'(0) = 0,$ and $\forall y\neq 0, f'(y) \neq 0.$
Therefore, $\forall y < 0, f'(y) < 0.$
Similarly, $\forall y > 0, f'(y) > 0.$
Therefore $f(y)$ must be strictly increasing on $(0,\infty)$
and $f(y)$ must be strictly decreasing on $(-\infty, 0).$
Therefore $y$ has a global minimum at $y = 0.$

Anyway...
As $y \to \infty, \log(3y^2 + 1) \to \infty.$

Therefore, as $y \to \infty ~f$ is unbounded.

user2661923
  • 35,619
  • 3
  • 17
  • 39
  • By letting f'(x) = 0, we get x = 4. This means we have *only* located a *relative* minimum (or maximum which tturned out to be not true). This does not prevent the function to have another value that is lower than the local minimum found at the endpoint(s) as indicated in your last statement. – Mick Oct 21 '20 at 08:00
  • I mean using the test of f'(x) = 0, followed by a verifying test (like first or second derivatives) can give us posiible *relative* extreme point.(s). As the function is continuous, we don't know for sure whether there is/are possible extreme point(s) at the far ends (when x is very positive (or very negative)). – Mick Oct 21 '20 at 16:11
  • I mean we cannot jump to that conclusion unless we have supporting arguement. The agruemnet that I can think of is "f'(x) > o when x > 4" implies f(x) is an increasing function and it has no x to make f(x) smaller than what we have found. – Mick Oct 21 '20 at 16:15
  • Since f(x) is an increasing function, for $x_n < x_{n + 1}, f(x_n) < f(x_{n+1})$ and the process never ends. This means no point on the right tail can call itself an extreme point. The same reasoning can be applied to the left tail. That is the only reason I can think of. – Mick Oct 22 '20 at 04:27
  • Plotting on Geogebra, I got a "$---_v---$" shaped graph.for f(x). – Mick Oct 22 '20 at 04:34
  • f(x) and f(y) are logically equivalent because one is displaced 4 units horizontally from the other. If, beside the claimed, no more points can be considered as extreme, the claimed one will then be the only minimum and is global. – Mick Oct 22 '20 at 04:51
  • yes, and it would be more complete if the part on decreasing/increasing arguement is included. We have to stop at this point. Thanks for your effort. – Mick Oct 22 '20 at 05:05
  • There are at least three typos in this answer. – PNDas Oct 22 '20 at 05:22
  • 1
    You are integrating $f(y)$ instead of $f'(y)$, next you wrote $y$ is increasing in $(0,inf)$,There is a similar typo in the next line. – PNDas Oct 22 '20 at 05:33
  • @PNDas +1 - thanks for your help - my blindness was working overtime, very careless answer. – user2661923 Oct 22 '20 at 05:44