# Math Help - Critical values and max/mins

1. ## Critical values and max/mins

This isn't quite homework, but still stuff I don't understand.

1- f(x) = 4x^2
0 < x ≤ 8

I understand that there's no local max, and that the absolute max is 256. But for local and absolute minimums I don't quite understand. How to find the min values. If I find the critical value and don't include 0, then there's no critical value, which means I don't understand how to find the mins.

Same thing with this one:

g(t) = | 4t - 9 |

g'(t) is 4. Which, as far as I know, doesn't mean there's a critical point.

One more question:

x^(-4)*ln(x)

Product rule. Right. But isn't the derivative of the left hand function x^(-4)*ln(-4), which doesn't exist because you can't take the log of a negative number?

2. Originally Posted by CFem
This isn't quite homework, but still stuff I don't understand.

1- f(x) = 4x^2
0 < x ≤ 8

I understand that there's no local max, and that the absolute max is 256. But for local and absolute minimums I don't quite understand. How to find the min values. If I find the critical value and don't include 0, then there's no critical value, which means I don't understand how to find the mins.

Same thing with this one:

g(t) = | 4t - 9 |

g'(t) is 4. Which, as far as I know, doesn't mean there's a critical point.

One more question:

x^(-4)*ln(x)

Product rule. Right. But isn't the derivative of the left hand function x^(-4)*ln(-4), which doesn't exist because you can't take the log of a negative number?
For the first, there's no absolute min.
Second, critical points are where $g' = 0$ or DNE. Here, it's at $t = \frac{9}{4}$. (BTW - $g' \ne 4$)

Third, not sure how you got the critical point of $x = -4$. The derivative is

$
\frac{1 - 4 \ln x}{x^5}
$

3. Originally Posted by Danny
For the first, there's no absolute min.
Second, critical points are where $g' = 0$ or DNE. Here, it's at $t = \frac{9}{4}$. (BTW - $g' \ne 4$)

Third, not sure how you got the critical point of $x = -4$. The derivative is

$
\frac{1 - 4 \ln x}{x^5}
$
For the first, does no absolute min imply that there's no local min?

What am I missing on the second derivative? The derivative of 4t is 4, the derivative of 9 is 0, no?

For the third one, could you show me your steps? I did this: (I was backwards in my first post)

F(x) = x^(-4)*ln(x)
f(x) = x^(-4)
f'(x) = -4x^(-5)
g(x) = ln(x)
g'(x) = 1/x

F'(x) = f'(x)g(x) + f(x)g'(x)
F'(x) = -4x^(-5) * ln(x) + $\frac{\frac{1}{x^4}}{x}$

4. Originally Posted by CFem
For the first, does no absolute min imply that there's no local min?

What am I missing on the second derivative? The derivative of 4t is 4, the derivative of 9 is 0, no?

For the third one, could you show me your steps? I did this: (I was backwards in my first post)

F(x) = x^(-4)*ln(x)
f(x) = x^(-4)
f'(x) = -4x^(-5)
g(x) = ln(x)
g'(x) = 1/x

F'(x) = f'(x)g(x) + f(x)g'(x)
F'(x) = -4x^(-5) * ln(x) + $\frac{\frac{1}{x^4}}{x}$
1) In general, abs. min doesn't imply there's no rel. min.
2) You have an absolute function

$g(t) = |4t-9| = \left\{\begin{array}{cl} 4t-9, & t \ge 9/4\\ -(4t-9),& t < 9/4\end{array}\right.$