# Given f(x) = 0.5x^{3} + x^{2} - x + 2, identify the x-intercepts and the points where the local...

## Question:

Given {eq}\displaystyle f(x) = 0.5x^{3} + x^{2} - x + 2 {/eq}, identify the {eq}x {/eq}-intercepts and the points where the local maximums and local minimums occur.

## Local Maxima and Minima

The local maxima of a function are the points within an interval at which the function is the maximum for all points within the interval. The local minima of a function are the points within an interval at which the function is the minimum for all points within the interval.

#### {eq}x {/eq}-intercepts

The {eq}x {/eq}-intercepts of a function {eq}f(x) {/eq} are the points where it is zero. So to find the x-intercepts, we need to find the roots of {eq}f(x) {/eq}.

Setting {eq}f(x) = 0 {/eq} yields:$$\displaystyle f(x) = 0.5x^{3} + x^{2} - x + 2 = 0, or\\ x^2 + 2x^2 -2x +4 =0$$

Using the polynomial roots calculator available at https://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php yields the roots of the equation as: {eq}x_1=-3.07395,\\ x_2=0.53697+1.00644i,\\ x_3=0.53697-1.00644i {/eq}

So the {eq}x-intercept {/eq} of {eq}f(x) {/eq} is -3.07395.

#### Local maxima and minima

To find the local maxima and minima, we need to set the first derivative of {eq}f(x) {/eq} to zero and solve for {eq}x {/eq}.

Setting the first derivative of {eq}f(x) {/eq} to zero yields:$$f'(x) = 1.5x^2 + 2x -1 = 0,\\ 3x^2 + 4x -2 = 0$$

Using the quadratic roots formula yields the extrema as: $$x_{1,2} = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{-4 \pm \sqrt{16+24}}{6} = \dfrac{\sqrt{10} \pm 2}{3} \approx (0,387, -1.721)$$

Next, we need to find which of these points are the local maxima and minima. For this, we need to check the value of the second derivative at the extrema. When the second derivative is negative, it is a local maximum and when it is positive, it is a local minimum.

Plugging in {eq}x_1 = 0.387, x_2 = -1.721 {/eq} yields:$$f''(x) = 3x +2,\\ f''(0.387) \approx 3.162 > 0 \Rightarrow x = 0.387 \text { is the local minimum},\\ f''(-1.721) \approx -3.163 < 0 \Rightarrow x = -1.721 \text { is the local maximum.}$$

Therefore the {eq}x {/eq}-intercept of {eq}\displaystyle f(x) = 0.5x^{3} + x^{2} - x + 2 {/eq} is -3.07395. The local maximum occurs at {eq}x = -1.721 {/eq} and the local minimum occurs at {eq}x = 0.387 {/eq}. 