MHB Can't Solve Character Schedule: f(x)=e^x(x^3-2x^2-x+4)-5

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The discussion revolves around discrepancies in the character schedule derived from the function f(x)=e^x(x^3-2x^2-x+4)-5. The original poster has correctly identified the roots but questions the character schedule provided in an old exam, noting differences in the factors. They suspect a potential typo in the exam's solution, specifically regarding the signs of the factors. Another participant emphasizes that math books can contain errors and suggests using computer algebra systems (CAS) to verify calculations. The conversation highlights the importance of double-checking mathematical work, especially when discrepancies arise.
Petrus
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Hello MHB,
I am working with an old exam that we got full soloution I got problem with their characters schedule ( hope you understand what I mean cause I could not find any translate)
1zvfjb.png

I got the same roots as the soloution but in the characters schedule why do they got $$x-3$$ and $$(x+1)^2$$ should it not be $$x+3$$ and $$(x-1)^2$$ or I am missing something, that's what I get when I solve it.
edit: the function maybe was not clearly to see in the picture but here it is $$f(x)=e^x(x^3-2x^2-x+4)-5$$
Regards,
$$|\pi\rangle$$
 
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I get $f'(x)=(x-1)^{2}(x+3)e^{x}$.
 
Petrus said:
...I got the same roots as the soloution but in the characters schedule why do they got $$x-3$$ and $$(x+1)^2$$ should it not be $$x+3$$ and $$(x-1)^2$$ or I am missing something, that's what I get when I solve it...

Post your working and we can get to the bottom of where you went wrong. :D
 
MarkFL said:
Post your working and we can get to the bottom of where you went wrong. :D
Hello Mark,
I get correct answer and graph but my characters schedule looks diffrent from their, I don't know how I can make one and when I make one in paint it does not look nice so I will describe with words. At left side of the characters schedule I got $$(x+3)$$ insted of $$(x-3)$$ and $$(x-1)^2$$ insted of $$(x+1)^2$$ and rest is exactly the same, so my guess is they made some sign typo? Just want to be sure so I don't make the characters schedule wrong. As Ackbach said I got the roots $$x=1$$ (<- double root) and $$x=-3$$.

Regards,
$$|\pi\rangle$$
 
Two comments:

1. You can make tables in $\LaTeX$ as follows:

Code:
\begin{array}{c|c|c}
   1 & 2 & 3 \\ \hline
   4 & 5 & 6 \\ \hline
   7 & f(x) & \sin(x) \\
\end{array}

produces

\begin{array}{c|c|c}
1 & 2 & 3 \\ \hline
4 & 5 & 6 \\ \hline
7 & f(x) & \sin(x) \\
\end{array}

2. Books can be wrong. Math books, especially calculus books, might have a lower error rate than some other types of books, but they definitely have incorrect things in them from time to time. In this case, there is no doubt that they computed the derivative incorrectly. In my opinion, in this day and age of CAS's, there's no excuse for not checking their work with a computer. CAS's can't replace a person, but people should use them to check their work!
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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