Question

Find the point on the curve y = x^3 − 11x + 5 at which the tangent is y = x − 11.

Answer 1

slope of y = x - 11 is 1

we know, slope of tangent of the curve y = x³ - 11x + 5 at (a,b) is 1st order derivatives of the curve at that given point.
hence, slope of tangent = $\bf{\frac{dy}{dx}|_{(a,b)}}$

y = x³ - 11x + 5
differentiate y with respect to x ,
dy/dx = 3x² - 11

Let (a, b) is the point on the curve y = x³ - 11x + 5 at which tangent is y = x - 11.
so, slope of tangent = dy/dx at (a,b) = 3a² - 11

now, slope of {y = x - 11} = slope of tangent
1 = 3a² - 11
=> 3a² = 12
=> a² = 4 => a = ±2

put a = -2 in b = a³ - 11a + 5 [(a,b) lies on y = x³ - 11x + 5 ]
b = (-2)³ - 11(-2) + 5 = -8 + 22 + 5 = 19

put a = 2 in b = a³ - 11a + 5
b = 2³ - 11(2) + 5 = 8 - 22 + 5 = -9

hence, required points are (-2,19) and (2,-9)

Answer 2

Answer:

Step-by-step explanation:

Hope you understood.