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

3 0

3 years ago

HELP PLS ASAP WILL GIVE 5 STARS

Andreyy89

Step-by-step explanation:

I do now know what this is about

7 0

2 years ago

Rewrite the product as a power:<br> <img src="https://tex.z-dn.net/?f=%28-a%29%5E%7B3%7D%20b%5E%7B3%7D" id="TexFormula1" title="

anastassius [24]

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${ - a}^{3} . {b}^{3} = ( { - a \times b})^{3} = ( { - ab})^{3} \\$

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4 0

3 years ago