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Problem 31 Hard Difficulty

Explain why Newton's method doesn't work for finding the root of the equation
$$ x^3 - 3x + 6 = 0 $$
if the initial approximation is chosen to be $ x_1 = 1 $.

Answer

$f(x)=x^{3}-3 x+6 \Rightarrow f^{\prime}(x)=3 x^{2}-3 .$ If $x_{1}=1,$ then $f^{\prime}\left(x_{1}\right)=0$ and the tangent line used for approximating $x_{2}$ is
horizontal. Attempting to find $x_{2}$ results in trying to divide by zero.

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Video Transcript

explain why Newton's method fails to find the roots of the equation. F of X equals zero, where FX is X cubed minus three x plus six. If we used the initial approximation, X one is equal to one. Answer this question. We should recall. Recall what Newton's method means geometrically. Let me just draw the graph of function and suppose that we want to find on approximation to this route here. What we would do is just choose a nearby point on the X axis and draw the tangent line to the graph of our function. At that point, no, we would do is see where the tangent line intersex the X axis again, and we'LL use that as our next approximation, and we'LL keep repeating this process until we are as close to the route as we like. So now if you write down the equation for this tangent lines, he were intersects the X axis. You can find the formula for the next approximation for Newton's method. So the way we get our next approximation as we take our current approximation except Ben and we subtract F of X have been divided by F prime of exhibition So to use this formula, say we want to find the next approximation, we would have to find the derivative. So let's compute the derivative. We use the power rule. We get that the derivative is three X squared minus three. All right, and now let's see what would happen to find X up to. We would need to compute f prime of X of one. So let's do that. We're given in the initial approximation is one. And now if we plug that in death prime, we just get three minus three, which is zero. Okay, and now we see where the problem is. We don't we're not able to find the next approximation because dividing by zero is not defined. So what's happening geometrically, let's say that here on the X axis is X one. And now, in this situation, since F prime of X one is zero, we have something that could possibly look like this. The point is that the tangent line at this point has slope zero, so it's horizontal, and that means that it will never cross the X axis again. So that's why we can't use this point to find the next approximation for Newton's method