https://file.aiquickdraw.com/mathbot/file/17266158391295qvcnqff.jpg

Let's go through the problem step by step based on the image you provided.

---

### Given: 
\[
f(t) = \frac{1}{t+1}
\]

---

### (a) Simplify the difference quotient:

The difference quotient is:

\[
\frac{f(t+h) - f(t)}{h}
\]

Substitute \( f(t) = \frac{1}{t+1} \) and \( f(t+h) = \frac{1}{t+h+1} \):

\[
\frac{f(t+h) - f(t)}{h} = \frac{\frac{1}{t+h+1} - \frac{1}{t+1}}{h}
\]

To simplify:

1. Find a common denominator for the difference in the numerator:
   \[
   \frac{1}{t+h+1} - \frac{1}{t+1} = \frac{(t+1) - (t+h+1)}{(t+h+1)(t+1)}
   \]
   Simplifying the numerator:
   \[
   (t+1) - (t+h+1) = -h
   \]
   Thus, we have:
   \[
   \frac{1}{t+h+1} - \frac{1}{t+1} = \frac{-h}{(t+h+1)(t+1)}
   \]

2. Now, substitute this back into the difference quotient:
   \[
   \frac{\frac{-h}{(t+h+1)(t+1)}}{h} = \frac{-1}{(t+h+1)(t+1)}
   \]
Thus, the simplified form is:
\[
\frac{-1}{(t+h+1)(t+1)}
\]

---

### (b) Calculate the derivative \( f'(t) \) using the limit definition:

We take the limit as \( h \to 0 \):

\[
f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h} = \lim_{h \to 0} \frac{-1}{(t+h+1)(t+1)}
\]

As \( h \to 0 \), \( t+h+1 \to t+1 \). So, the derivative is:

\[
f'(t) = \frac{-1}{(t+1)^2}
\]

---

### (c) Evaluate \( f'(1) \):

Substitute \( t = 1 \) into the expression for \( f'(t) \):

\[
f'(1) = \frac{-1}{(1+1)^2} = \frac{-1}{4}
\]

---

### (d) Find the equation of the tangent line at \( (1, f(1)) \):

First, evaluate \( f(1) \):

\[
f(1) = \frac{1}{1+1} = \frac{1}{2}
\]

The point of tangency is \( (1, \frac{1}{2}) \), and the slope of the tangent line is \( f'(1) = \frac{-1}{4} \).

Using the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
where \( m = \frac{-1}{4} \), \( (x_1, y_1) = (1, \frac{1}{2}) \).

Substitute these values:
\[
y - \frac{1}{2} = \frac{-1}{4}(x - 1)
\]

Simplify the equation:

\[
y = \frac{-1}{4}(x - 1) + \frac{1}{2}
\]
\[
y = \frac{-1}{4}x + \frac{1}{4} + \frac{1}{2}
\]
\[
y = \frac{-1}{4}x + \frac{3}{4}
\]

Thus, the equation of the tangent line is:
\[
y = \frac{-1}{4}x + \frac{3}{4}
\]



Let f(t) = 1/(t + 1). (a) Simplify the following difference quotient as much as possible: f(t + h) - f(t) / h. (b) Use your result from (a) and the limit definition of the derivative to calculate f'(t). (c) Use your answer from part (b) to evaluate f'(1). (d) Find the equation of the tangent line to the curve at the point (1, f(1)).

Explore the process of finding derivatives using the limit definition for the function f(t) = 1/(t + 1), including simplifying the difference quotient and calculating the equation of the tangent line at a specific point. Suitable for college students studying calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation