Function Values E99Ff7

1. **State the problem:** We are given the function $$F(b) = \frac{b - b^2}{1 + b^2}$$ and need to find the values of $F(0)$, $F(1)$, $F(\frac{1}{2})$, and $F(\tan x)$. 2. **Recall the formula:** $$F(b) = \frac{b - b^2}{1 + b^2}$$ 3. **Calculate $F(0)$:** Substitute $b=0$ into the formula: $$F(0) = \frac{0 - 0^2}{1 + 0^2} = \frac{0}{1} = 0$$ 4. **Calculate $F(1)$:** Substitute $b=1$: $$F(1) = \frac{1 - 1^2}{1 + 1^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$$ 5. **Calculate $F(\frac{1}{2})$:** Substitute $b=\frac{1}{2}$: $$F\left(\frac{1}{2}\right) = \frac{\frac{1}{2} - \left(\frac{1}{2}\right)^2}{1 + \left(\frac{1}{2}\right)^2} = \frac{\frac{1}{2} - \frac{1}{4}}{1 + \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{5}{4}}$$ Simplify the fraction: $$= \frac{\frac{1}{4}}{\frac{5}{4}} = \frac{1}{4} \times \frac{4}{5} = \frac{1}{5}$$ 6. **Calculate $F(\tan x)$:** Substitute $b=\tan x$: $$F(\tan x) = \frac{\tan x - (\tan x)^2}{1 + (\tan x)^2} = \frac{\tan x - \tan^2 x}{1 + \tan^2 x}$$ Recall the identity: $$1 + \tan^2 x = \sec^2 x$$ Rewrite numerator: $$\tan x - \tan^2 x = \tan x (1 - \tan x)$$ So: $$F(\tan x) = \frac{\tan x (1 - \tan x)}{\sec^2 x} = \tan x (1 - \tan x) \cos^2 x$$ **Final answers:** $$F(0) = 0$$ $$F(1) = 0$$ $$F\left(\frac{1}{2}\right) = \frac{1}{5}$$ $$F(\tan x) = \tan x (1 - \tan x) \cos^2 x$$