MHB Differentiation help (stationary points)

[Lhh]

I’m struggling with questions c, e and f.
I don’t think I understand how to find stationary points.

 

[skeeter]

stationary points occur where the derivative is zero or is undefined.
in this case, $L’$ is defined everywhere in the function’s given domain.

$L(\lambda) = \lambda^{150}e^{-3\lambda}$

$L’(\lambda) = 150 \lambda^{149} e^{-3\lambda} - 3\lambda^{150} e^{-3\lambda}$

$L’(\lambda) = 3\lambda^{149}e^{-3\lambda} (50-\lambda)$

$L’(\lambda) = 0$ at $\lambda = 50$

$\lambda <50 \implies L’ > 0 \implies L$ is increasing.
$\lambda >50 \implies L’ <0 \implies L$ is decreasing
therefore, $L$ has a maximum at $\lambda =50$
this method is called the 1st derivative test for extrema.

You could also evaluate the value of $L’’(50)$ to determine if the stationary point is a maximum or minimum.
if $L’’(50) <0$, then $L(50)$ is a maximum
if $L’’(50) >0$, then $L(50)$ is a minimum
this is the 2nd derivative test for extrema.

see what you can get done with the log function

 

[HOI]

stationary points occur where the derivative is zero or is undefined.
in this case, $L’$ is defined everywhere in the function’s given domain.
$L’(\lambda) = 0$ at $\lambda = 50$

$L’(\lambda) = 0$ at $\lambda = 50$ OR at $\lambda= 0$

$\lambda <50 \implies L’ > 0 \implies L$ is increasing.
$\lambda >50 \implies L’ <0 \implies L$ is decreasing
therefore, $L$ has a maximum at $\lambda =50$
this method is called the 1st derivative test for extrema.

You could also evaluate the value of $L’’(50)$ to determine if the stationary point is a maximum or minimum.
if $L’’(50) <0$, then $L(50)$ is a maximum
if $L’’(50) >0$, then $L(50)$ is a minimum
this is the 2nd derivative test for extrema.

see what you can get done with the log function

 

[skeeter]

$L’(\lambda) = 0$ at $\lambda = 50$ OR at $\lambda= 0$

$\lambda = 0$ is not in the given domain of the function ... reference the problem statement in post #1