the graph of y=e^tanx-2 crosses the x axis at one point in the interval [0,1]. What is the slope of the graph at this point? I g et 3.08267 but it's not the right answer
1 answer:
Y = e^tanx - 2 To find at which point it crosses x axis we state that y= 0 e^tanx - 2 = 0 e^tanx = 2 tanx = ln 2 tanx = 0.69314 x = 0.6061 to find slope at that point first we need to find first derivative of funtion y. y' = (e^tanx)*1/cos^2(x) now we express x = 0.6061 in y' and we get: y' = k = 2,9599
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