Point J has coordinates (10,8). The midpoint of the line segment JK is (6,12).
1 answer:
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Answer:
(a)
(b)
Step-by-step explanation:
The given function is
.... (1)
According to the first principle of the derivative,
Cancel out common factors.
By applying limit, we get
.... (2)
Therefore .
(b)
Put x=2, to find the y-coordinate of point of tangency.
The coordinates of point of tangency are (2,0.25).
The slope of tangent at x=2 is
Substitute x=2 in equation 2.
The slope of the tangent line at x=2 is -0.25.
The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).
Using point slope form the equation of tangent is
Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.
The root 
can be converted into the power 
. Therefore we can rewrite the problem as 
and then follow the exponent rules about a power to a power, multiplying 1/2 and 3/4 together.
Thus the problem becomes
, which then can be turned into ![\sqrt[8]{10} ^{3x}](https://tex.z-dn.net/?f=%20%5Csqrt%5B8%5D%7B10%7D%20%5E%7B3x%7D)
, making the last option our answer.
Thus the problem becomes
