Question

Find the slope of the tangent to the curve y = x^3 − 3x + 2 at the point whose x-coordinate is 3.

Answer 1

given curve , y = x³ - 3x + 2
we have to find slope of tangent of the curve at the point whose x - co-ordinate is 3.e.g., x = 3

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

so, Let's differentiate with respect to x,
dy/dx = d(x³ - 3x + 2)/dx
= 3x² - 3
at x = 3, dy/dx = 3(3)² - 3 = 3 × 9 - 3 =27 - 3 = 24
hence, slope of tangent of the curve is 24 at x = 3

Answer 2

Step-by-step explanation:

given curve , y = x³ - 3x + 2

we have to find slope of tangent of the curve at the point whose x - co-ordinate is 3.e.g., x = 3

we know, slope of tangent of curve y = f(x) at a point (a,b) is 1st order derivatives at that given point. e.g., \textbf{slope of tangent}=\frac{dy}{dx}|_{(a,b)}slope of tangent=

dx

dy

∣

(a,b)

so, Let's differentiate with respect to x,

dy/dx = d(x³ - 3x + 2)/dx

= 3x² - 3

at x = 3, dy/dx = 3(3)² - 3 = 3 × 9 - 3 =27 - 3 = 24

hence, slope of tangent of the curve is 24 at x = 3