1. **Problem statement:** Find the function $f(x)$ given that the slope of the tangent line to $f$ at $x=c$ is given by the limit $$\lim_{h \to 0} \frac{\tan\left(\frac{\pi}{4} + h\right) - 1}{h}.$$ Also find the value of $c$. 2. **Understanding the limit:** The limit expression is the definition of the derivative of $f$ at $x=c$: $$f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}.$$ Here, the expression inside the limit matches the derivative definition with $f(c+h) = \tan\left(\frac{\pi}{4} + h\right)$ and $f(c) = 1$. 3. **Identify $f(x)$:** Since $f(c+h) = \tan\left(\frac{\pi}{4} + h\right)$, it suggests that $f(x) = \tan(x)$ shifted so that $x = c$ corresponds to $\frac{\pi}{4}$. 4. **Find $c$:** We know $f(c) = 1$ and $f(x) = \tan(x)$, so: $$f(c) = \tan(c) = 1.$$ The angle where $\tan(c) = 1$ in the principal interval is: $$c = \frac{\pi}{4}.$$ 5. **Confirm the derivative:** The derivative of $f(x) = \tan(x)$ is: $$f'(x) = \sec^2(x).$$ At $x = \frac{\pi}{4}$: $$f'\left(\frac{\pi}{4}\right) = \sec^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\cos\left(\frac{\pi}{4}\right)}\right)^2 = \left(\frac{1}{\frac{\sqrt{2}}{2}}\right)^2 = 2.$$ **Final answers:** - $f(x) = \tan(x)$ - $c = \frac{\pi}{4}$