Question

use estimation use a graphing calculator to estimate the x-coordinates at which any extrema occur for the given function. round to the nearest hundredth. if there is no relative maximum or no relative minimum, write none. ( f(x) = x^3 + 3x^2 - 6x - 6 ) relative maxima ( x = ) blank relative minima ( x = ) blank need help with this question? get a hint

Explanation:

Step1: Find the derivative of \( f(x) \)

The function is \( f(x) = x^3 + 3x^2 - 6x - 8 \). The derivative \( f'(x) \) is found using the power rule. For \( x^n \), the derivative is \( nx^{n - 1} \). So, \( f'(x)=3x^{2}+6x - 6 \).

Step2: Set the derivative equal to zero and solve for \( x \)

To find critical points, set \( f'(x) = 0 \):
\[
3x^{2}+6x - 6 = 0
\]
Divide both sides by 3:
\[
x^{2}+2x - 2 = 0
\]
Use the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) for \( ax^{2}+bx + c = 0 \). Here, \( a = 1 \), \( b = 2 \), \( c=-2 \).
\[
x=\frac{-2\pm\sqrt{2^{2}-4(1)(-2)}}{2(1)}=\frac{-2\pm\sqrt{4 + 8}}{2}=\frac{-2\pm\sqrt{12}}{2}=\frac{-2\pm2\sqrt{3}}{2}=-1\pm\sqrt{3}
\]
Calculate the numerical values:
\( \sqrt{3}\approx1.732 \), so \( x=-1 + 1.732\approx0.73 \) and \( x=-1 - 1.732\approx - 2.73 \)

Step3: Determine relative maxima and minima (using second derivative or graph)

The second derivative \( f''(x)=6x + 6 \).

• For \( x\approx - 2.73 \), \( f''(-2.73)=6(-2.73)+6=6(-1.73)\approx - 10.38<0 \), so there is a relative maximum at \( x\approx - 2.73 \).
• For \( x\approx0.73 \), \( f''(0.73)=6(0.73)+6=4.38 + 6 = 10.38>0 \), so there is a relative minimum at \( x\approx0.73 \).

Answer:

Relative maximum: \( x\approx - 2.73 \)
Relative minimum: \( x\approx0.73 \)