Finding Local Extrema for the Polynomial Function

To find the points where the function $f(x) = 2x^3-15x^2 +36x+10$ has a local maximum and a local minimum, we need to use calculus, specifically the first and second derivatives.

Step 1: Calculate the First Derivative

The first derivative of the function, denoted as $f'(x)$, tells us the slope of the tangent line at any point $x$. Local maxima and minima occur where the slope is zero, meaning $f'(x)=0$.

Given function: $f(x) = 2x^3-15x^2 +36x+10$

Calculating the first derivative using the power rule ($\frac{d}{dx}(ax^n) = anx^{n-1}$): $f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(15x^2) + \frac{d}{dx}(36x) + \frac{d}{dx}(10)$ $f'(x) = 2(3x^{3-1}) - 15(2x^{2-1}) + 36(1x^{1-1}) + 0$ $f'(x) = 6x^2 - 30x + 36$

Step 2: Find Critical Points

Critical points are the values of $x$ where the first derivative is either zero or undefined. Since $f'(x)$ is a polynomial, it is defined for all real numbers. Therefore, we only need to find where $f'(x)=0$.

Set the first derivative to zero: $6x^2 - 30x + 36 = 0$

To simplify, divide the entire equation by 6: $\frac{6x^2}{6} - \frac{30x}{6} + \frac{36}{6} = \frac{0}{6}$ $x^2 - 5x + 6 = 0$

Factor the quadratic equation: We look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. $(x-2)(x-3) = 0$

This gives us two critical points: $x-2 = 0 \implies x = 2$ $x-3 = 0 \implies x = 3$

So, the critical points are $x=2$ and $x=3$. These are the potential locations for local maxima or minima.

Step 3: Use the Second Derivative Test

The second derivative test helps determine whether a critical point corresponds to a local maximum or minimum. We calculate the second derivative, $f''(x)$, and evaluate it at the critical points.

Calculate the second derivative by differentiating $f'(x)$: $f'(x) = 6x^2 - 30x + 36$ $f''(x) = \frac{d}{dx}(6x^2) - \frac{d}{dx}(30x) + \frac{d}{dx}(36)$ $f''(x) = 6(2x^{2-1}) - 30(1x^{1-1}) + 0$ $f''(x) = 12x - 30$

Now, evaluate $f''(x)$ at the critical points:

• At $x=2$: $f''(2) = 12(2) - 30 = 24 - 30 = -6$ Since $f''(2) < 0$, the function has a local maximum at $x=2$.
• At $x=3$: $f''(3) = 12(3) - 30 = 36 - 30 = 6$ Since $f''(3) > 0$, the function has a local minimum at $x=3$.

Conclusion

The local maximum occurs at $x=2$, and the local minimum occurs at $x=3$. This matches option 3.