Question

1. (5 points) if $y(x) = \sin(x)$, find $y^{(4)}(x)$.
2. (15 points) find points on the curve where the tangent line is horizontal for $y = 2x^3 + 3x^2 - 12x + 9$.

Explanation:

Step1: Recall Horizontal Tangent Condition

A horizontal tangent line has a slope of 0. The slope of the tangent to a curve \( y = f(x) \) is given by the derivative \( f'(x) \). So, we need to find \( f'(x) \) and set it equal to 0, then solve for \( x \), and finally find the corresponding \( y \)-values.

Step2: Find the Derivative of the Function

Given \( y = 2x^3 + 3x^2 - 12x + 9 \), we use the power rule for differentiation, which states that if \( f(x) = ax^n \), then \( f'(x) = nax^{n - 1} \).

Applying the power rule term - by - term:

• The derivative of \( 2x^3 \) is \( 2\times3x^{3 - 1}=6x^2 \)
• The derivative of \( 3x^2 \) is \( 3\times2x^{2 - 1}=6x \)
• The derivative of \( - 12x \) is \( - 12\times1x^{1 - 1}=-12 \)
• The derivative of the constant 9 is 0.

So, \( y'=f'(x)=6x^2 + 6x-12 \)

Step3: Set the Derivative Equal to Zero and Solve for \( x \)

We set \( f'(x) = 0 \):

$$
LATEXBLOCK0
$$

Using the zero - product property (if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \)):

• If \( x+2 = 0 \), then \( x=-2 \)
• If \( x - 1=0 \), then \( x = 1 \)

Step4: Find the Corresponding \( y \)-Values

We substitute \( x=-2 \) and \( x = 1 \) into the original function \( y = 2x^3+3x^2-12x + 9 \)

For \( x=-2 \):

$$
LATEXBLOCK1
$$

For \( x = 1 \):

$$
LATEXBLOCK2
$$

Answer:

The points on the curve where the tangent line is horizontal are \((-2,29)\) and \((1,2)\)