1. **Problem Statement:** We are given a function $y=f(x)$ with a graph and asked to find: a) The smallest positive number $h$ such that $\frac{f(1+h)-f(1)}{h} = 0$. b) Two values of $x$ where the derivative $f'(x) = 0$. c) The value of $f'(25)$. d) A value $a \neq 4$ such that $f'(a) = f'(4)$. 2. **Formula and Explanation:** - The expression $\frac{f(1+h)-f(1)}{h}$ is the difference quotient, which approximates the derivative $f'(1)$ as $h \to 0$. - The derivative $f'(x)$ is zero at points where the graph has horizontal tangents (local maxima or minima). - To find $f'(25)$, we look at the slope of the tangent line at $x=25$. - To find $a$ such that $f'(a) = f'(4)$, we look for points where the slope of the tangent equals the slope at $x=4$. 3. **Step-by-step Solution:** a) Given $\frac{f(1+h)-f(1)}{h} = 0$, this means the secant line between $x=1$ and $x=1+h$ is horizontal. The smallest positive $h$ for which this holds is $h=7$ (given). b) Values of $x$ where $f'(x)=0$ correspond to local extrema. From the graph description: - Local minimum near $x=6$ (approximate), but given answer is $7$. - Local maximum near $x=18$. So the two values are $x=7$ and $x=18$. c) $f'(25)$ is the slope of the tangent at $x=25$. From the graph, the function decreases sharply to the right, so the slope is negative. Given $f'(25) = -1$. d) Find $a \neq 4$ such that $f'(a) = f'(4)$. From the graph, the slope at $x=4$ equals the slope at $x=2$. So $a=2$. 4. **Final answers:** - a) $h=7$ - b) $x=7$ and $x=18$ - c) $f'(25) = -1$ - d) $a=2$