# Thread: Finding the roots of an equation with its derivative?

1. ## Finding the roots of an equation with its derivative?

I remember my friend showing me a way to find the max/min points of an equation with its derivative and remember him faintly telling me you can also find the roots with its derivative. I was wondering if this were true and if there are any faster methods than the factor/remainder theorem (guess and check + long division).

2. You can by solving $\dfrac{dy}{dx} = 0$ to find a stationary point. You find if it's a min or max by checking the sign of $\dfrac{d^2y}{dx^2}$

3. Originally Posted by e^(i*pi)
You can by solving $\dfrac{dy}{dx} = 0$ to find a stationary point. You find if it's a min or max by checking the sign of $\dfrac{d^2y}{dx^2}$
Thanks, but I was wondering if you could find the roots (zeros) of a function.

4. Originally Posted by youngb11
Thanks, but I was wondering if you could find the roots (zeros) of a function.
depends on the function ... could you be more specific?

5. Originally Posted by skeeter
depends on the function ... could you be more specific?
Say $F(x)=x^3-4x^2+x+6$

Derivative would be $3x^2-8x+1$. Would I be able to use that to figure out the roots/zeros of a function?

6. Originally Posted by youngb11
Say $F(x)=x^3-4x^2+x+6$

Derivative would be $3x^2-8x+1$. Would I be able to use that to figure out the roots/zeros of a function?
since $b^2-4ac = 52$ , the quadratic will not factor over the rational numbers ... use the quadratic formula.

$\displaystyle x = \frac{4\pm \sqrt{13}}{3}$

7. It looks to me like both e^(i pi) and skeeter have misunderstood the question. e^(i pi) talked about finding critical points and skeeter about solving f'(x)= 0. The question, I believe, is about using f'(x) to solve f(x)= 0.

It sounds to me like you are talking about "Newton's method" for solving equations. The basic idea is that you replace the function with its tangent line approximation and solve the resulting linear equation. Of course, a given graph will have different tangent lines at different points on its graph so you have to select a starting value for x. If we start with $x_0$ as the starting value, the slope of the tangent line is $f '(x_0)$ and the tangent line approximation to f(x), about the point $(x_0, y_0)$, is $y= f '(x_0)(x- x_0)+ f(x_0)$. Setting that equal to 0, we have $f '(x_0)(x- x_0)+ f(x_0)= 0$ so $f '(x_0)(x- x_0)= -f(x_0)$, $x- x_0= -\frac{f(x_0)}{f '(x_0)}$, and, finally, $x= x_0- \frac{f(x_0)}{f'(x_0)}$. As long as you stay away from values of $x_0$ that make $f '(x_0)$ close to 0 (so that fraction will not "blow up") that x will be reasonably close to the solution to f(x)= 0. You can make it better by repeating this- choose " $x_0$" to be this new value of x and repeat- as often as you like.

Taking your example, let $F(x)= x^3- 4x^2+ x+ 6$ so that $F'(x)= 3x^2- 8x+ 1$. Taking $x_0= 1$, because that is close to the min point of the quadratic so away from F'= 0, we have F(1)= 1- 4+ 1+ 6= 4, F '(1)= 3- 8+ 1= -4 and $x= 1- (4/-4)= 2$. Now, F(2)= 8- 16+ 2+ 6= 0 and F'(2)= 24- 16+ 1= 9 so that $x= 2- (0/9)= 2$. Of course, that is because F(2)= 0 so that 2 is, in fact, a solution.

Suppose the equation had been $f(x)= x^2- 2$ (so that we are trying to find the square root of 2). Now $f'= 2x$. Taking $x_= 1$, we have f(1)= 1- 2= -1 and f'(1)= 2. Then $x= 1- (-1/2)= 3/2= 1.5$. Taking $x_0= 3/2$ now, f(3/2)= 9/4- 8/4= 1/4 and f'(3/2)= 3. $x= 3/2- ((1/4)/3)= 3/2- 1/12= 18/12- 1/12= 17/12= 1.41666.$.

As you can see that is converging pretty rapidly to $\sqrt{2}= 1.414....$.