1. **Problem Statement:** We are given a function $f$ defined on the interval $[a,b]$ with $f(b) > f(a)$ and $a \leq x \leq b$. The derivative $f'(x)$ exists for all $x$ in $(a,b)$ except at $x=0$. We want to find the number of values $c$ in $(a,b)$ such that $$\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \frac{f(b) - f(a)}{b - a}.$$ This limit is the definition of the derivative at $c$, so we are looking for points where the instantaneous rate of change equals the average rate of change over $[a,b]$. 2. **Relevant Theorem:** This is a direct application of the **Mean Value Theorem (MVT)** which states: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c \in (a,b)$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a}.$$ 3. **Important Notes:** - The function $f$ is continuous on $[a,b]$. - The derivative exists everywhere on $(a,b)$ except at $x=0$. - The MVT requires differentiability on the open interval, but here $f'(0)$ does not exist. 4. **Implication of the derivative not existing at $x=0$:** Since $f'(0)$ does not exist, the MVT does not guarantee a single $c$ but the problem asks for the number of $c$ where the derivative (or the limit defining it) equals the average rate of change. 5. **Analyzing the graph and behavior:** - The graph dips below the x-axis near zero, indicating a cusp or corner at $x=0$ where derivative fails. - The function has a local maximum and a local minimum in $(a,b)$. - The average slope $\frac{f(b)-f(a)}{b-a}$ is positive since $f(b) > f(a)$. 6. **Number of points where the slope equals the average slope:** - The graph crosses the average slope line multiple times. - Because of the cusp at $x=0$, the function can have multiple points where the secant slope equals the instantaneous slope. - By the shape described, there are **four** such points. 7. **Conclusion:** The number of values $c$ in $(a,b)$ such that $$\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \frac{f(b) - f(a)}{b - a}$$ is **four**. **Final answer:** (D) Four