Finding Approximate Critical Number with Newton's Method

[PirateCat]

Use Newton's Method to approximate a critical number of the function \(\displaystyle \displaystyle f(x) = \frac{1}{2} x^8 + \frac{6}{5} x ^5+ 2 x +10\) near the point \(\displaystyle x = 2 \). Use \(\displaystyle x_1 = 2 \) as the initial approximation. Find the next two approximations, x_2 and x_3, to four decimal places each.I have been working on this problem and made f(x) a derivative and then use 2 as the initial approximation, but the math website keeps telling me it is wrong. I do not understand what it means by "critical number" when it comes to the Newton's Method. Any help would be greatly appreciated, thank you!

 

[I like Serena]

Use Newton's Method to approximate a critical number of the function \(\displaystyle \displaystyle f(x) = \frac{1}{2} x^8 + \frac{6}{5} x ^5+ 2 x +10\) near the point \(\displaystyle x = 2 \). Use \(\displaystyle x_1 = 2 \) as the initial approximation. Find the next two approximations, x_2 and x_3, to four decimal places each.I have been working on this problem and made f(x) a derivative and then use 2 as the initial approximation, but the math website keeps telling me it is wrong. I do not understand what it means by "critical number" when it comes to the Newton's Method. Any help would be greatly appreciated, thank you!

Hi PirateCat! Welcome to MHB! ;)

I'm assuming that a critical number is the same thing as a critical point.
That is where the derivative is zero.

However, Newton's method (actually the Newton-Raphson method) finds a zero.
Can it be that we're supposed to take the derivative and then find the zero?
\begin{tikzpicture}[>=stealth, xscale=5, font=\large, scale=0.5]
\foreach \i in {-1,0,1,2} {%
\draw (\i,.1) -- (\i,-.1) node[below] {$\i$};%
}
\foreach \i in {-5,0,5,10,15} {%
\draw (.1,\i) -- (-.1,\i) node

{$\i$};%
}
\draw[->] (-1.7,0) -- (2.2,0) node

{$x$};
\draw[->] (0,-6) -- (0,16) node[above] {$y$};
\draw[domain=-1.5:1.1, smooth, variable=\x, ultra thick, blue] plot ({\x},{1/2 * (\x)^8 + 6/5*(\x)^5 + 2*(\x) + 10}) node

{$y=f(x)$};
\draw[domain=-1.3:1.0, smooth, variable=\x, ultra thick, red] plot ({\x},{4 * (\x)^7 + 6*(\x)^4 + 2}) node

{$y=f'(x)$};
\end{tikzpicture}

Which approximations did you find? (Wondering)​

 

[PirateCat]

The approximations that I got were around 1.7024 and 1.4022 if I remember correctly; they were around that range. That makes sense equaling it out to zero, we did not go over that in class though.

 

[I like Serena]

The approximations that I got were around 1.7024 and 1.4022 if I remember correctly; they were around that range. That makes sense equaling it out to zero, we did not go over that in class though.

It seems to be intended as an example where Newton-Raphson has some limitations.
The approximations won't quite converge, since we have to go up from x=2, before going down to the actual critical point. (Nerd)

 

[PirateCat]

I see, that makes sense. Would linear approximation have to come into play then?

 

[I like Serena]

I see, that makes sense. Would linear approximation have to come into play then?

The Newton-Raphson method is a linear approximation! (Thinking)

 

[PirateCat]

Ahhh yeah you're right. I forgot that you can use the L(x) equation and turn it into Newton's Method XD I'm still confused on where to start though for this problem.

 

[I like Serena]

Ahhh yeah you're right. I forgot that you can use the L(x) equation and turn it into Newton's Method XD I'm still confused on where to start though for this problem.

What do you mean by "starting"?
I'm guessing we're supposed to find successive approximations, which will be a bit weird since we won't immediately go to the actual critical point.
Typically we'll overshoot after which we'll converge after all.