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)=x25e2xf(x)=\frac{x^2}{5e^{2x}}f(x)=5e2xx2 In the interval [−12,2][-\frac{1}{2},2][−21,2] Final Answer Show final answer x=−12,x=0x=-\frac{1}{2}, x=0x=−21,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 – Extremum to a polynomial function in a closed interval – Exercise 6872 July 28, 2019 Extremum, Increase and Decrease Sections – A rational function – Exercise 6820 July 24, 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 – A polynomial – Exercise 6805 July 24, 2019 Extremum, Increase and Decrease sections – Min/Max problems (maximal area) – Exercise 6884 July 28, 2019 Extremum, Increase and Decrease sections – Extremum to a polynomial function in a closed interval – Exercise 6876 July 28, 2019 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ
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