On what interval is f(x) = ln(3x) concave down?

[calculus 1983]

** f(x) is concave down where f''(x) < 0 **

f(x) = ln(3x) = 1 / 3x x 3 = 1/x = x ^ -1

f''(x) = -1x ^ -2 = 1/x^2 set it < 0

I need to find where f''(x) is negative and since the numerator is negative (-) the bottom has to be positive, which it will be because anything squared becomes positive (+) ... so a - / + = a negative (-)

The final answer is (0,00) but i don't understand why it is (0,00)
can anyone explain to me why that is the final answer? Thank you.

 

[daon]

f''(x) = -1x ^ -2 = 1/x^2 set it < 0

Problem there.. where'd your negative go?

f''(x) = -1/x^2, right?

Well f''(x) is then always negative. Since 1/x^2 > 0 for all x, -1/x^2 < 0 for all x. See?

So for any x at all, your function will be concave down. But for x <= 0 your original function is undefined. So it will be concave down for (0, \(\displaystyle \infty\)).

 

[calculus 1983]

ah i see ... i had f''(x) = -1 / x^2 set < 0 but i forgot to put the negative sign in front of the one ...which is probably what i get for trying to study for calc at 1 in the morning lol

 

[daon]

ah i see ... i had f''(x) = -1 / x^2 set < 0 but i forgot to put the negative sign in front of the one ...which is probably what i get for trying to study for calc at 1 in the morning lol

This is the best time for math! Less distractions and the power of coffee to propel you. Plus there's the famous quote by Adrian Mathesis: "All great theorems were discovered after midnight."

Good luck.

 

[calculus 1983]

"So for any x at all, your function will be concave down. But for x <= 0 your original function is undefined. So it will be concave down for (0, oo)."

Wait, why is it that for any x at all, your function will be concave down? Because no matter what you put for x, whether negative or positive, the end result becomes negative?

 

[daon]

"So for any x at all, your function will be concave down. But for x <= 0 your original function is undefined. So it will be concave down for (0, oo)."

Wait, why is it that for any x at all, your function will be concave down? Because no matter what you put for x, whether negative or positive, the end result becomes negative?

What I meant was, that since the second derivative is negative for all real numbers, your function should be concave down for all x... BUT your original function is not defined for \(\displaystyle \(- \infty, o\]\).. it would not make sense to say that f(x) is concave down at -4.. right? So for all x except those x in that interval it will be concave down. Alternatively, could say "f(x) is always concave down" or "it is concave down for all x in the domain of f(x)" since the answer is precisely that.

 

[calculus 1983]

Do you always go back to your original function to check? .. because for instance, i did another problem similar to this one which asked on what interval is g(x) = 1 / x^2 + 1 concave down? ..... and eventually i got g''(x) = 6x^2 - 2 / (x^2 +1) ^2 ...and since the denominator is always going to turn out negative we needed to set 6x^2 - 2 (the numerator) < 0

add 2 on both sides .... 6x^2 < 2 divide 6 on both sides ...take the square root on each side leaving you with x < + or minus radical 1/3
and the final answer was (- radical 1/3 , + radical 1/3)
and i did not go back to g(x) = 1 / x^2 + 1 (the original function)

 

[daon]

add 2 on both sides .... 6x^2 < 2 divide 6 on both sides ...take the square root on each side leaving you with x < + or minus radical 1/3
and the final answer was (- radical 1/3 , + radical 1/3)
and i did not go back to g(x) = 1 / x^2 + 1 (the original function)

It didn't matter in this case since g(x) was defined for all x. Your f(x)=ln(3x) is only defined on \(\displaystyle \( 0, \infty \)\). It wouldn't make sense to say that ln(3x) is concave down at 0 or a negative number. You should always take into consideration the domain of your original function.