Question

Find points at which the tangent to the curve y = x^3 − 3x^2 − 9x + 7 is parallel to the x-axis.

Answer 1

given curve , y = x³ - 3x² - 9x + 7
we have to find the point at which the tangent to the curve is parallel to the x - axis.

we know, slope of tangent of curve , y = f(x) at (a,b) is 1st order derivatives at that point (a,b)
hence, slope of tangent = $\bf{\frac{dy}{dx}|_{(a,b)}}$

y = x³ - 3x² - 9x + 7
differentiate y with respect to x,
dy/dx = 3x² - 6x - 9
= 3(x² - 2x - 3)
= 3(x² - 3x + x - 3)
= 3(x - 3)(x + 1)

Let at (a,b) , y is parallel to the x - axis.
then, slope of tangent = 0
so, dy/dx = 3(x - 3)(x + 1) = 0
x = 3 , -1

put x = 3 in the curve y = x³ - 3x² - 9x + 7
y = (3)³ - 3(3)² - 9(3) + 7
y = 3³ - 3³ - 27 + 7 = -20

put x = -1 , in the curve y = x³ - 3x² - 9x + 7
y = (-1)³ - 3(-1)² - 9(-1) + 7
= -1 - 3 + 9 + 7 = 5 + 7 = 12

hence, (3, -20) and (-1, 12) are the point at which the tangent to the curve, y = x³ - 3x² - 9x + 7 is parallel to the axis.