Question #48346
Give the equation of the line tangent to the curve f(x) = (x^(2) + 1)/(e^(2x)+sqrt(x)) at the point (1,f(1))
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#340153 on Dec 2023
#340153 on Dec 2023
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Dear Bob. Thank you for adding information.
Dear Bender. Thank you for adding information.
Shouldn't the derivative of e^(2x) be 2e^(2x) and wouldn't m = (2(e^2+1) - 2(2e^2+(1/2)) / (e^2+1)^2) = (2e^2+2 - 4e^2 -1) / (e^2+1)^2 m = -2e^2+1/(e^2+1)^2 m= approx -.1958 Then sub m for = -.1958, y = 2/(e^2+1), and x = 1 to solve for b b= approx 0.4342
Shouldn't the derivative of e^(2x) be 2e^(2x) and wouldn't m = (2(e^2+1) - 2(2e^2+(1/2)) / (e^2+1)^2) = (2e^2+2 - 4e^2 -1) / (e^2+1)^2 m = (-2e^2+1) / (e^2+1)^2 m= approx -.1958 Then sub m for = -.1958, y = 2/(e^2+1), and x = 1 to solve for b b= approx 0.4342