1. **State the problem:** We have an exponential function $f(x) = ab^x$ and two points: $f(-3.5) = 22$ and $f(2) = 64$. We need to find $f(1)$ to the nearest tenth. 2. **Write the equations from the points:** $$ab^{-3.5} = 22$$ $$ab^{2} = 64$$ 3. **Divide the second equation by the first to eliminate $a$:** $$\frac{ab^{2}}{ab^{-3.5}} = \frac{64}{22}$$ $$b^{2 - (-3.5)} = \frac{64}{22}$$ $$b^{5.5} = \frac{64}{22}$$ 4. **Solve for $b$:** $$b = \left(\frac{64}{22}\right)^{\frac{1}{5.5}}$$ 5. **Calculate $b$ approximately:** $$b \approx \left(2.9091\right)^{0.1818} \approx 1.224$$ 6. **Substitute $b$ back into one of the original equations to find $a$:** Using $ab^{-3.5} = 22$: $$a = \frac{22}{b^{-3.5}} = 22 \times b^{3.5}$$ $$a = 22 \times (1.224)^{3.5}$$ 7. **Calculate $a$ approximately:** $$a \approx 22 \times 2.011 = 44.242$$ 8. **Find $f(1)$:** $$f(1) = ab^{1} = 44.242 \times 1.224 = 54.1$$ **Final answer:** $$f(1) \approx 54.1$$