find the points on the curve ( y = x^{3}+3x^{2}-9x + 4 ) where the tangent is horizontal.\nsmaller ( x…

Explanation:

Step1: Find the derivative of the function

The derivative of (y = x^{3}+3x^{2}-9x + 4) using the power rule ((x^{n})^\prime=nx^{n - 1}) is (y^\prime=3x^{2}+6x-9).

Step2: Set the derivative equal to zero

Since the tangent is horizontal when (y^\prime = 0), we solve (3x^{2}+6x - 9=0). Divide through by (3) to get (x^{2}+2x-3 = 0). Factor the quadratic equation: ((x + 3)(x - 1)=0).

Step3: Solve for (x)

Using the zero - product property (a\cdot b=0) implies (a = 0) or (b = 0). If (x+3=0), then (x=-3). If (x - 1=0), then (x = 1).

Step4: Find the corresponding (y) - values

When (x=-3): (y=(-3)^{3}+3(-3)^{2}-9(-3)+4=-27 + 27+27 + 4=31). When (x = 1): (y=(1)^{3}+3(1)^{2}-9(1)+4=1 + 3-9 + 4=-1).

Answer:

smaller (x) - value ((x,y)=(-3,31)) larger (x) - value ((x,y)=(1,-1))