Answer:
Is this an exam? looks like it
Inflection point is the point where the second derivative of a graph is zero.
y = (x+1)arctan xy' = (x+1)(arctan x)' + (1)arctan xy' = (x+1)/(x^2+1) + arctan xy'' = (x+1)(1/(1+x^2))' + 1/(1+x^2) + 1/(1+x^2)y'' = (x+1)(-1/(1+x^2)^2)(2x)+2/(1+x^2)y'' = ((x+1)(-2x)+1+x^2)/(1+x^2)^2y'' = (-2x^2-2x+2+2x^2)/(1+x^2)^2y'' = (-2x+2)/(1+x^2)^2
Solving for point of inflection: y'' = 00 = (-2x+2)/(1+x^2)^20 = -2x+2x = 1y(1) = (1+1)arctan(1) = 2 * pi/4 = pi/2
Therefore, E(1, pi/2).
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Answer:
h = 1.403733341286
Step-by-step explanation:
17.
the triangle of sides ‘l’ and ‘l-h’ is a right angled triangle.
Then
Then
Then
18.
For l = 6 and θ = 40 :
= 6-6cos(40)
= 1.403733341286
