1. **Problem statement:** Use the Intermediate Value Theorem (IVT) to verify that $f(x)$ has a zero in the given interval, then use the bisection method to find an interval of length $\frac{1}{32}$ containing the zero for $f(x) = x^2 - 7$ on $[2,3]$. 2. **Intermediate Value Theorem:** If $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists $c \in (a,b)$ such that $f(c) = 0$. 3. **Check signs:** $$f(2) = 2^2 - 7 = 4 - 7 = -3 < 0$$ $$f(3) = 3^2 - 7 = 9 - 7 = 2 > 0$$ Since $f(2) < 0$ and $f(3) > 0$, by IVT there is a zero in $(2,3)$. 4. **Bisection method:** Start with interval $[2,3]$ length 1. - Midpoint $m_1 = \frac{2+3}{2} = 2.5$ - Evaluate $f(2.5) = 2.5^2 - 7 = 6.25 - 7 = -0.75 < 0$ - Since $f(2.5) < 0$ and $f(3) > 0$, zero lies in $[2.5,3]$ length $0.5$ 5. Repeat bisection to reduce interval length to $\frac{1}{32} = 0.03125$: - $[2.5,3]$, midpoint $m_2 = 2.75$, $f(2.75) = 7.5625 - 7 = 0.5625 > 0$ - Zero in $[2.5,2.75]$, length $0.25$ - $[2.5,2.75]$, midpoint $m_3 = 2.625$, $f(2.625) = 6.890625 - 7 = -0.109375 < 0$ - Zero in $[2.625,2.75]$, length $0.125$ - $[2.625,2.75]$, midpoint $m_4 = 2.6875$, $f(2.6875) = 7.22265625 - 7 = 0.22265625 > 0$ - Zero in $[2.625,2.6875]$, length $0.0625$ - $[2.625,2.6875]$, midpoint $m_5 = 2.65625$, $f(2.65625) = 7.05419921875 - 7 = 0.05419921875 > 0$ - Zero in $[2.625,2.65625]$, length $0.03125$ 6. The interval $[2.625,2.65625]$ has length $\frac{1}{32}$ and contains the zero. **Final answer:** Zero lies in $[2.625,2.65625]$.