Hello I am doing some problems from Calculus, in the Newton's method, and it says to compute the $\min|f'(x)|\ \ (-1 <= x <= 1)$
My $f(x)$ is $x^3/2 - 2x^2 + 3x +2$, then my $f'(x)$ is $3x^2/2 - 4x + 3$
But I do not understand what this "min" is and how to I have to find it? ($x$ can be from $[-1, 1]$)
On
I'm going to assume your book means $\min(\vert f'(x)\vert: x \in [-1,1])$.
This means the lowest value of $f'(x)$ that you can get if you choose $x$ in the given interval. Finding the minimum of a function with calculus takes a few steps -- have you seen the "first derivative test" before? You are going to want to use it here to minimize $f'(x)$. First you need to find the points where $\frac{d}{dx} f'(x) = 0$. Your minimum will be at one of these points, or at one of the endpoints of the interval, so you need to check the value of $f'(x)$ at all of them. Do these steps make sense?
$f'(x)$ is a quadratic polynomial in $x$, and the extremum of a quadratic polynomial $ax^2+bx+c$ is attained at $x=-\dfrac b{2a}$.
Its value is $\dfrac{4ac-b^2}{4a}$, and it is a minimum if $a>0$, a maximum if $a<0$.
Now this minimum is here attained at $x=\dfrac43>1$, so the function is decreasing on the interval $[-1,1]$, so the minimum on this interval is attained at $x=1$.