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Find an equation for the tangent line to the curve y = 2*√x at the given point ( 1, 2). Show your work

User Guy
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Answer:

y = x+1

Explanation:

The slope of the tangent line is the value of the derivative at the given point:

y' = 2(1/2x^(-1/2)) = 1/√x

For x=1, this value is ...

y' = 1/√1 = 1

The point-slope form of the equation for a line is ...

y -k = m(x -h) . . . . . . line with slope m through point (h, k)

For m = 1 and (h, k) = (1, 2) the equation of the line is ...

y -2 = 1(x -1)

y -2 = x -1 . . . . point-slope equation, simplified a bit

y = x +1 . . . . add 2 to get slope-intercept form

_____

Additional comment

We used the "power rule" for finding the derivative. That rule tells you ...

for y = x^a, the derivative is y' = a(x^(a-1))

In this problem, the exponent 'a' is 1/2, so (a-1) is -1/2.

Find an equation for the tangent line to the curve y = 2*√x at the given point ( 1, 2). Show-example-1
User Fefe
by
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