Question

given the function, $f(x) = x^3 - 3x - 2$, what are the coordinates of the relative minimum? ( , )

Explanation:

Step1: Find the first derivative

To find critical points, we first find the derivative of \( f(x) = x^3 - 3x - 2 \). Using the power rule, the derivative \( f'(x) = 3x^2 - 3 \).

Step2: Find critical points

Set \( f'(x) = 0 \) to find critical points:

$$ 3x^2 - 3 = 0 $$

Factor out 3:

$$ 3(x^2 - 1) = 0 $$

Which simplifies to:

$$ x^2 - 1 = 0 \implies (x - 1)(x + 1) = 0 $$

So the critical points are \( x = 1 \) and \( x = -1 \).

Step3: Find the second derivative

To determine if these critical points are minima or maxima, we find the second derivative. Differentiating \( f'(x) = 3x^2 - 3 \), we get \( f''(x) = 6x \).

Step4: Test critical points with second derivative

• For \( x = 1 \): \( f''(1) = 6(1) = 6 > 0 \), so the function is concave up at \( x = 1 \), meaning there is a relative minimum at \( x = 1 \).
• For \( x = -1 \): \( f''(-1) = 6(-1) = -6 < 0 \), so the function is concave down at \( x = -1 \), meaning there is a relative maximum at \( x = -1 \).

Step5: Find the y - coordinate of the relative minimum

Now we find the value of \( f(x) \) at \( x = 1 \). Substitute \( x = 1 \) into \( f(x) \):

$$ f(1)=1^3 - 3(1)-2=1 - 3 - 2=-4 $$

Answer:

\((1, -4)\)