Q. 185.0( 1 Vote )

# Find the point on

Given; y = x3 - 11x + 5 and the equation of tangent is y = x - 11

Slope of the tangent is the coefficient of x [y = mx + c] = 1 From the first equation; From (i) and (ii)

3x2 − 11 = 1

3x2 = 12

x = ±2

y = x − 11 = −13, −9

The required points are (2,−9) and (−2,−13)

OR

Given; √49.5

Let f(x) = √x and f(x + δx) = √(x + δx)

Here; x = 49 and δx = 0.5

We know; f(x + δx) = δx f(x) + f(x)

Here;   = 7.036

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