Question

1. find all relative maxima and minima. indicate which, if any, are absolute maxima or minima. ( y = 2x^3 - 3x^2 - 12x + 8 )

Explanation:

Step 1: Find the first derivative

To find relative maxima and minima, we first find the derivative of the function \( y = 2x^{3}-3x^{2}-12x + 8 \). Using the power rule \( \frac{d}{dx}(x^{n})=nx^{n - 1} \), we get:
\( y'=\frac{d}{dx}(2x^{3}-3x^{2}-12x + 8)=6x^{2}-6x - 12 \)

Step 2: Find critical points

Set the first derivative equal to zero and solve for \( x \):
\( 6x^{2}-6x - 12 = 0 \)
Divide both sides by 6:
\( x^{2}-x - 2=0 \)
Factor the quadratic equation:
\( (x - 2)(x + 1)=0 \)
So the critical points are \( x = 2 \) and \( x=- 1 \)

Step 3: Find the second derivative

To determine if these critical points are maxima or minima, we find the second derivative. Differentiate \( y' = 6x^{2}-6x - 12 \):
\( y''=\frac{d}{dx}(6x^{2}-6x - 12)=12x-6 \)

Step 4: Test critical points using the second derivative test

• For \( x=-1 \):

Substitute \( x = - 1 \) into \( y'' \):
\( y''(-1)=12(-1)-6=-12 - 6=-18<0 \)
Since \( y''(-1)<0 \), the function has a relative maximum at \( x=-1 \)
To find the \( y \)-value of the relative maximum, substitute \( x = - 1 \) into the original function:
\( y(-1)=2(-1)^{3}-3(-1)^{2}-12(-1)+8=2(-1)-3(1)+12 + 8=-2-3 + 12 + 8 = 15 \)

• For \( x = 2 \):

Substitute \( x = 2 \) into \( y'' \):
\( y''(2)=12(2)-6=24 - 6 = 18>0 \)
Since \( y''(2)>0 \), the function has a relative minimum at \( x = 2 \)
To find the \( y \)-value of the relative minimum, substitute \( x = 2 \) into the original function:
\( y(2)=2(2)^{3}-3(2)^{2}-12(2)+8=2(8)-3(4)-24 + 8=16-12-24 + 8=-12 \)

Step 5: Analyze for absolute maxima and minima

The function \( y = 2x^{3}-3x^{2}-12x + 8 \) is a cubic function with a positive leading coefficient (\( 2>0 \)). As \( x
ightarrow\infty \), \( y
ightarrow\infty \) and as \( x
ightarrow-\infty \), \( y
ightarrow-\infty \). So there is no absolute maximum or absolute minimum.

Answer:

Relative maximum: \( (-1,15) \) (no absolute maximum)
Relative minimum: \( (2, - 12) \) (no absolute minimum)