Use the definition of the instantaneous rate of change to the expression for the slope of the tangent line to find the function 2 square root x plus 3 at any point 

To find the slope of the tangent line to the function \( f(x) = 2\sqrt{x} + 3 \) at any point \( x = a \), we use the definition of the instantaneous rate of change, which is the derivative of the function at that point. The derivative gives us the slope of the tangent line.

### Step 1: Definition of the Derivative
The derivative of \( f(x) \) at a point \( x = a \) is defined as:

\[
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}
\]

### Step 2: Substitute the Function into the Definition
Substitute \( f(x) = 2\sqrt{x} + 3 \) into the definition:

\[
f'(a) = \lim_{h \to 0} \frac{(2\sqrt{a+h} + 3) - (2\sqrt{a} + 3)}{h}
\]

### Step 3: Simplify the Expression
Simplify the expression by canceling the constant terms (the "+3" cancels out):

\[
f'(a) = \lim_{h \to 0} \frac{2\sqrt{a+h} - 2\sqrt{a}}{h}
\]

Factor out the common factor of 2 in the numerator:

\[
f'(a) = \lim_{h \to 0} \frac{2(\sqrt{a+h} - \sqrt{a})}{h}
\]

### Step 4: Rationalize the Numerator
To simplify this limit, multiply the numerator and the denominator by the conjugate of the numerator \( \sqrt{a+h} + \sqrt{a} \):

\[
f'(a) = \lim_{h \to 0} \frac{2(\sqrt{a+h} - \sqrt{a})(\sqrt{a+h} + \sqrt{a})}{h(\sqrt{a+h} + \sqrt{a})}
\]

The numerator becomes a difference of squares:

\[
f'(a) = \lim_{h \to 0} \frac{2((a+h) - a)}{h(\sqrt{a+h} + \sqrt{a})}
\]

Simplify the expression:

\[
f'(a) = \lim_{h \to 0} \frac{2h}{h(\sqrt{a+h} + \sqrt{a})}
\]

Cancel the \( h \) terms:

\[
f'(a) = \lim_{h \to 0} \frac{2}{\sqrt{a+h} + \sqrt{a}}
\]

### Step 5: Take the Limit as \( h \to 0 \)
As \( h \) approaches 0, \( \sqrt{a+h} \) approaches \( \sqrt{a} \):

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

### Conclusion
The slope of the tangent line to the function \( f(x) = 2\sqrt{x} + 3 \) at any point \( x = a \) is:

\[
f'(a) = \frac{1}{\sqrt{a}}
\]

This result gives you the instantaneous rate of change of the function at any point \( x = a \).

Would you like more details on any specific step, or do you have any questions?

Here are some related questions:

1. How would you find the equation of the tangent line at a specific point \( x = a \)?
2. What does the slope of the tangent line represent in the context of the function?
3. How can the concept of the derivative be applied to other types of functions?
4. What is the physical interpretation of the derivative in real-world problems?
5. How does the derivative relate to the concept of a rate of change in various fields of study?

**Tip:** When dealing with square root functions, rationalizing the numerator can be a helpful technique to simplify limits.

Learn how to find the slope of the tangent line to the function f(x) = 2 sqrt(x) + 3 at any point using the definition of the derivative. This problem covers calculus concepts including the instantaneous rate of change and the derivative. Suitable for college-level students and those 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