Demier

To find the derivative of \( f(x) = \frac{1}{x+1} \) at \( x = 3 \) using the limit definition and then determine the equation of the tangent line at that point, follow these steps: ### Step 1: Apply the Limit Definition of the Derivative The derivative of \( f \) at \( x = a \) is: \[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \] For \( a = 3 \): \[ f'(3) = \lim_{h \to 0} \frac{f(3+h) - f(3)}{h} \] Calculate \( f(3+h) \) and \( f(3) \): \[ f(3+h) = \frac{1}{(3+h)+1} = \frac{1}{4+h}, \quad f(3) = \frac{1}{4} \] Substitute into the limit: \[ f'(3) = \lim_{h \to 0} \frac{\frac{1}{4+h} - \frac{1}{4}}{h} \] Combine the fractions in the numerator: \[ \frac{1}{4+h} - \frac{1}{4} = \frac{4 - (4+h)}{4(4+h)} = \frac{-h}{4(4+h)} \] Now divide by \( h \): \[ f'(3) = \lim_{h \to 0} \frac{-h}{4(4+h)h} = \lim_{h \to 0} \frac{-1}{4(4+h)} \] Evaluate the limit as \( h \to 0 \): \[ f'(3) = \frac{-1}{4(4+0)} = -\frac{1}{16} \] ### Step 2: Find the Equation of the Tangent Line Using the point-slope form \( y = f(a) + f'(a)(x - a) \): - \( a = 3 \), \( f(3) = \frac{1}{4} \), and \( f'(3) = -\frac{1}{16} \). Substitute these values: \[ y = \frac{1}{4} - \frac{1}{16}(x - 3) \] Simplify: \[ y = -\frac{1}{16}x + \frac{3}{16} + \frac{4}{16} = -\frac{1}{16}x + \frac{7}{16} \] ### Final Answer **Derivative at \( x = 3 \):** \[ \boxed{-\frac{1}{16}} \] **Equation of the Tangent Line:** \[ \boxed{y = -\frac{1}{16}x + \frac{7}{16}} \]

I'm gonna get a prince husband! *I hug my pillow*

*He chuckles, finding your enthusiasm endearing* Well, you'll have to be patient. Princes aren't just going to fall out of the sky, you know.

I pout. *a loud thud comes from outside*

*He notices and sighs, trying to soften his demeanor* I'm not going to hurt you. I just need to know your name, and how old you are.

*she let's out a soft yip, she doesn't know how to talk yet*

*He notices the yip, and his expression softens even more* You're just a baby. I see. I have a question, how old are you? Can you answer it?