Determine all the extrema of $y(x)=(5x-1)e^{4x}$.
$y(-\frac{1}{20})=-1\frac{1}{4}e^{-\frac{1}{5}}$ is a minimum
There are no extrema.
$y(\frac{1}{5})=0$ is a maximum.
The correct answer is not among the other options.
Determine all the extrema of $y(x)=(5x-1)e^{4x}$.
Antwoord 1 correct
Correct
Antwoord 2 optie
There are no extrema.
Antwoord 2 correct
Fout
Antwoord 3 optie
$y(\frac{1}{5})=0$ is a maximum.
Antwoord 3 correct
Fout
Antwoord 4 optie
The correct answer is not among the other options.
Antwoord 4 correct
Fout
Antwoord 1 optie
$y(-\frac{1}{20})=-1\frac{1}{4}e^{-\frac{1}{5}}$ is a minimum
Antwoord 1 feedback
Correct: $y'(x)=(20x+1)e^{4x}$. Hence, the stationary point is $x=-\frac{1}{20}$. $y''(x)=(80x+24)e^{4x}$. Consequently, $y''(-\frac{1}{20})=20e^{-\frac{1}{5}}>0$ and hence, $y(-\frac{1}{20})=-1\frac{1}{4}e^{-\frac{1}{5}}$ is a minimum.

Go on.
Antwoord 2 feedback
Wrong: $y'(x)=5e^{4x}+(5x-1)\cdot e^{4x} \cdot 4$.

See Chain rule.
Antwoord 3 feedback
Wrong: $y'(x)=5e^{4x}+(5x-1)\cdot e^{4x} \cdot 4$.

See Chain rule.
Antwoord 4 feedback
Wrong: The correct answer is among them.

Try again.