find the equation of the line tangent to…?

y= (3)(cuberoot of (6x+2)) – tan (3x-3+ pi/4) at x=1

please show the work because im soo confused!

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A few days ago

Eric212

Favorite Answer

plug x=1 into the equation and what do you get for y?

You know the line has to go through that point.

y=6-tan(pi/4)=6-1=5

If the line is tangent at (1,5), it is parallel – or has the same slope.

I’m assuming you know how to take derivatives (slopes).

y’ = (6x+2)^-(2/3) – 3sec2(3x-3+pi/4)

at x=1

y’ = (8)^-(2/3) – 3sec2(pi/4)

y’ = 0.25 – 6 = -5.75

whats the line that passes through the point (1,5) with a slope of -5.75?

y-5=-5.75(x-1)

y-5=-5.75x+5.75

y=-5.75x-10.75

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