Extremum, Increase and Decrease Sections – Calculate absolute minimum and maximum in a closed interval – Exercise 5488 Post category:Extremum, Increase and Decrease Sections Post comments:0 Comments Exercise Find the minimum and maximum points of the function f(x)=\frac{x^2}{5e^{2x}} In the interval [-\frac{1}{2},2] Final Answer Show final answer x=-\frac{1}{2}, x=0 Solution Coming soon… Share with Friends Read more articles Previous PostExtremum, Increase and Decrease Sections – Proof of inequality – Exercise 2208 Next PostExtremum, Increase and Decrease Sections – A polynomial – Exercise 6805 You Might Also Like Extremum, Increase and Decrease Sections – x multiplied by an exponential function – Exercise 6831 July 25, 2019 Extremum, Increase and Decrease sections – Min/Max problems (maximal volume) – Exercise 6897 July 29, 2019 Extremum, Increase and Decrease sections – Extremum to an exponential function in a closed interval – Exercise 6911 July 29, 2019 Extremum, Increase and Decrease sections – Extremum to a polynomial function in a closed interval – Exercise 6872 July 28, 2019 Extremum, Increase and Decrease Sections – A polynomial – Exercise 6814 July 24, 2019 Extremum, Increase and Decrease sections – Min/Max problems (maximal slope) – Exercise 6893 July 29, 2019 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ
Extremum, Increase and Decrease Sections – x multiplied by an exponential function – Exercise 6831 July 25, 2019
Extremum, Increase and Decrease sections – Min/Max problems (maximal volume) – Exercise 6897 July 29, 2019
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Extremum, Increase and Decrease sections – Extremum to a polynomial function in a closed interval – Exercise 6872 July 28, 2019
Extremum, Increase and Decrease sections – Min/Max problems (maximal slope) – Exercise 6893 July 29, 2019