Solving F(x)=-2x^3+5x^2+4x-3 Find Roots And Turning Points
In this comprehensive guide, we delve into the intricacies of solving cubic equations and determining the turning points of a given function. Our focus will be on the specific function $f(x)=-2x3+5x2+4x-3$. We will explore the methods for finding the roots of the equation $f(x)=0$ and calculating the coordinates of the turning points, which are crucial in understanding the behavior and characteristics of the cubic function. This article aims to provide a step-by-step approach, making it easier for students and enthusiasts to grasp the underlying concepts and techniques. The solution of cubic equations and finding turning points are fundamental concepts in calculus and algebra, with applications spanning various fields, including physics, engineering, and economics.
2.3.1 Solving for $x$ if $f(x) = 0$
To solve for x when $f(x) = 0$, we need to find the roots of the cubic equation $-2x^3 + 5x^2 + 4x - 3 = 0$. Solving cubic equations can be more challenging than solving quadratic equations, as there isn't a straightforward formula like the quadratic formula that directly gives the roots. However, several methods can be employed to find the solutions, including factoring, using the Rational Root Theorem, and numerical methods. The Rational Root Theorem is particularly useful for identifying potential rational roots, which can then be verified through synthetic division or direct substitution. Factoring, if possible, simplifies the equation into a product of lower-degree polynomials, making it easier to find the roots. Numerical methods, such as the Newton-Raphson method, provide approximate solutions when analytical methods are difficult to apply. In this section, we will explore these methods in detail and apply them to our given function. First, we can try to factor the cubic equation by looking for rational roots. By the Rational Root Theorem, any rational root of the polynomial must be of the form $\frac{p}{q}$, where $p$ is a factor of the constant term (-3) and $q$ is a factor of the leading coefficient (-2). The possible rational roots are $\pm 1$, $\pm 3$, $\pm \frac{1}{2}$, and $\pm \frac{3}{2}$. We can test these values by substituting them into the function $f(x)$. If $f(a) = 0$, then $x = a$ is a root of the equation. Let's test $x = 1$: $f(1) = -2(1)^3 + 5(1)^2 + 4(1) - 3 = -2 + 5 + 4 - 3 = 4 \neq 0$. So, $x = 1$ is not a root. Next, let's try $x = -1$: $f(-1) = -2(-1)^3 + 5(-1)^2 + 4(-1) - 3 = 2 + 5 - 4 - 3 = 0$. Thus, $x = -1$ is a root of the equation. Since $x = -1$ is a root, $(x + 1)$ must be a factor of the polynomial. We can perform polynomial long division or synthetic division to find the other factor. Using synthetic division:
-1 | -2 5 4 -3
2 -7 3
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-2 7 -3 0
This gives us the quotient $-2x^2 + 7x - 3$. So, we can rewrite the cubic equation as: $(x + 1)(-2x^2 + 7x - 3) = 0$ Now, we need to solve the quadratic equation $-2x^2 + 7x - 3 = 0$. We can use the quadratic formula: $x = \frac-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = -2$, $b = 7$, and $c = -3$. Substituting these values, we get}2(-2)}$ $x = \frac{-7 \pm \sqrt{49 - 24}}{-4}$ $x = \frac{-7 \pm \sqrt{25}}{-4}$ $x = \frac{-7 \pm 5}{-4}$ This gives us two solutions{-4} = \frac{-2}{-4} = \frac{1}{2}$ $x_2 = \frac{-7 - 5}{-4} = \frac{-12}{-4} = 3$ Therefore, the roots of the cubic equation $-2x^3 + 5x^2 + 4x - 3 = 0$ are $x = -1$, $x = \frac{1}{2}$, and $x = 3$. These are the x-values for which the function $f(x)$ equals zero.
2.3.2 Calculating the Coordinates of B and E, the Turning Points of $f$
To calculate the coordinates of the turning points of $f(x) = -2x^3 + 5x^2 + 4x - 3$, we need to find the points where the derivative of the function, $f'(x)$, is equal to zero. These points are also known as critical points. The turning points represent local maxima or minima of the function, where the function changes its direction from increasing to decreasing or vice versa. Finding the turning points is a crucial step in analyzing the behavior of a function and sketching its graph. The derivative of a function gives the slope of the tangent line at any point on the function's curve. At the turning points, the tangent line is horizontal, and hence the derivative is zero. In this section, we will first find the derivative of the given function and then solve for the points where the derivative is zero. After finding the x-coordinates of the turning points, we will substitute these values back into the original function to find the corresponding y-coordinates. This will give us the exact coordinates of the turning points, which are essential for understanding the function's graph and behavior. First, let's find the derivative of $f(x)$: $f'(x) = \fracd}{dx}(-2x^3 + 5x^2 + 4x - 3)$ Using the power rule for differentiation, we get}2a}$ where $a = 3$, $b = -5$, and $c = -2$. Substituting these values, we get}2(3)}$ $x = \frac{5 \pm \sqrt{25 + 24}}{6}$ $x = \frac{5 \pm \sqrt{49}}{6}$ $x = \frac{5 \pm 7}{6}$ This gives us two solutions6} = \frac{12}{6} = 2$ $x_2 = \frac{5 - 7}{6} = \frac{-2}{6} = -\frac{1}{3}$ So, the x-coordinates of the turning points are $x = 2$ and $x = -\frac{1}{3}$. Now, we need to find the corresponding y-coordinates by substituting these values back into the original function $f(x)$3}$3}) = -2(-\frac{1}{3})^3 + 5(-\frac{1}{3})^2 + 4(-\frac{1}{3}) - 3$ $f(-\frac{1}{3}) = -2(-\frac{1}{27}) + 5(\frac{1}{9}) - \frac{4}{3} - 3$ $f(-\frac{1}{3}) = \frac{2}{27} + \frac{5}{9} - \frac{4}{3} - 3$ To combine these fractions, we need a common denominator, which is 27{3}) = \frac{2}{27} + \frac{15}{27} - \frac{36}{27} - \frac{81}{27}$ $f(-\frac{1}{3}) = \frac{2 + 15 - 36 - 81}{27}$ $f(-\frac{1}{3}) = \frac{-100}{27}$ So, the other turning point is $(-\frac{1}{3}, -\frac{100}{27})$. Therefore, the coordinates of the turning points B and E are $(2, 9)$ and $(-\frac{1}{3}, -\frac{100}{27})$. These points are crucial for sketching the graph of the cubic function, as they indicate where the function changes direction. To determine whether each turning point is a local maximum or minimum, we can use the second derivative test or analyze the sign of the first derivative around these points. The turning points’ coordinates provide valuable information about the function's behavior and are essential in various applications of calculus.
In summary, we have successfully solved the cubic equation $f(x) = -2x^3 + 5x^2 + 4x - 3 = 0$ and found its roots to be $x = -1$, $x = \frac{1}{2}$, and $x = 3$. Additionally, we calculated the coordinates of the turning points of the function $f(x)$, which are $(2, 9)$ and $(-\frac{1}{3}, -\frac{100}{27})$. These solutions and coordinates provide a comprehensive understanding of the function's behavior and characteristics. The methods used in this guide, such as the Rational Root Theorem, synthetic division, the quadratic formula, and differentiation, are fundamental tools in algebra and calculus. Mastering these techniques is essential for solving more complex problems and gaining a deeper understanding of mathematical functions. The ability to solve cubic equations and find turning points has wide-ranging applications in various fields, making these concepts an integral part of mathematical education. By following the step-by-step approach outlined in this article, students and enthusiasts can confidently tackle similar problems and enhance their mathematical skills.
