# Use the bisection method to find the minimum of the function

Use the bisection method to find the minimum of the function is $f(x)= 3x^2–4x+1$ over the interval $[0,2]$ . Determine the optimal value of $x$ within $5\%$ of the initial interval and How many function evaluations are needed to get within x* ± 0.001?

My attempt

The first iteration $a=0,b=2, m=(a+b)/2=1$

$f(a)=1, f(b)=5, f(m)=0$

$x^*=6x-4=0$

$x^*=2/3=0.6667$

I do not understand what the first part question is asking.Is it mean that I need to continue the evaluation until I find the value of $b-a$ is closed to $0.05$? After the 11 evaluations , I get a=0.3321 and b=0.3341.But x*=0.6667 instead of 0.3333

• I think you need to think about the criterion by which you decide whether to take the left-hand interval, or the right-hand interval at each step. Normally, bisection is used to find roots of continuous functions, where, by using the Intermediate Value Theorem, you just choose the interval such that the function changes sign on that interval. Using calculus, or even algebra, we can see that the minimum occurs at $x=\dfrac23$. – Adrian Keister Apr 17 '18 at 15:46