2) Given f(x) = 2x³ - 7x? – 3x + 8a) Find all roots of the function and simplify them as much as possibleb) Sketch the complete graphc) indicate the viewing windowd) Find the x-interceptse) Find the relative maximum and minimum to the nearest hundredth

the given function is fx=2x3-7x2-3x+8 find f1 f1=213-712-31+8=2-7-3+8=10-10=0 therefore, 1 is one of…

Answered: 2) Given f(x) = 2x' - 7x² – 3x + 8 a)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Concept explainers

Question

2) Given f(x) = 2x³ - 7x? – 3x + 8
a) Find all roots of the function and simplify them as much as possible
b) Sketch the complete graph
c) indicate the viewing window
d) Find the x-intercepts
e) Find the relative maximum and minimum to the nearest hundredth

Expert Solution

Step 1

the given function is $f (x) = 2 x^{3} - 7 x^{2} - 3 x + 8$

find $f (1)$

$\begin{array}{rcl} f (1) & = & 2 {(1)}^{3} - 7 {(1)}^{2} - 3 (1) + 8 \\ = & 2 - 7 - 3 + 8 \\ = & 10 - 10 \\ = & 0 \end{array}$

therefore, 1 is one of the root of the given function

to find the other roots perform the synthetic division

$2 - 7 - 3 8 0 2 - 5 - 8 1 2 - 5 - 8 0$

Step 2

therefore, $2 x^{3} - 7 x^{2} - 3 x + 8 = (x - 1) (2 x^{2} - 5 x + 8)$

$2 x^{3} - 7 x^{2} - 3 x + 8 = (x - 1) (2 x^{2} - 5 x + 8)$

find the value of x from $(2 x^{2} - 5 x + 8)$ using the quadratic formula to get other roots

$\begin{array}{rcl} x & = & \frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 (2) (- 8)}}{2 (2)} \\ = & \frac{5 \pm \sqrt{25 + 64}}{2 (2)} \\ = & \frac{5 \pm \sqrt{89}}{4} \\ = & \frac{5 + \sqrt{89}}{4}, \frac{5 - \sqrt{89}}{4} \end{array}$

therefore, the roots of the given function are $\begin{array}{rcl} x & = & 1, \frac{5 + \sqrt{89}}{4}, \frac{5 - \sqrt{89}}{4} \end{array}$

Step by step

Solved in 3 steps with 1 images

Knowledge Booster

$Parabolas$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions