1. **Problem Statement:** We have three functions defined as: $$f(x) = 2x - 1, \quad g(x) = 3x + 2, \quad h(x) = 5^x.$$ We need to find: (i) $f\left(\frac{1}{2}\right)$ (ii) $h(0)$ (iii) $g(-3)$ 2. **Step (i): Find $f\left(\frac{1}{2}\right)$** - Substitute $x = \frac{1}{2}$ into $f(x)$: $$f\left(\frac{1}{2}\right) = 2 \times \frac{1}{2} - 1$$ - Simplify: $$= 1 - 1 = 0$$ 3. **Step (ii): Find $h(0)$** - Substitute $x = 0$ into $h(x)$: $$h(0) = 5^0$$ - Recall any nonzero number to the power 0 is 1: $$= 1$$ 4. **Step (iii): Find $g(-3)$** - Substitute $x = -3$ into $g(x)$: $$g(-3) = 3 \times (-3) + 2$$ - Simplify: $$= -9 + 2 = -7$$ 5. **Step (c)(i): Find $g'(x)$** - $g(x) = 3x + 2$ is a linear function. - The derivative of $g(x)$ with respect to $x$ is the coefficient of $x$: $$g'(x) = 3$$ 6. **Step (c)(ii): Find $x$ when $g'(x) = 4$** - Since $g'(x) = 3$ for all $x$, it never equals 4. - Therefore, there is no value of $x$ such that $g'(x) = 4$. **Final answers:** (i) $f\left(\frac{1}{2}\right) = 0$ (ii) $h(0) = 1$ (iii) $g(-3) = -7$ (c)(i) $g'(x) = 3$ (c)(ii) No solution since $g'(x)$ is constant 3 and never equals 4.