the slope of the tangent line to f(x) at x=a can be found by evaluating the limit f'(a)=mtan=limx->a f(x)-f(a)/x-a provided the limit exists. use this limit to find f'(-3) given f(x)=1/x-3. find f'(-3)=


To find \( f'(-3) \) for the function \( f(x) = \frac{1}{x - 3} \), we will use the definition of the derivative:

\[
f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}
\]

In this case, \( a = -3 \), so the expression for \( f'(-3) \) becomes:

\[
f'(-3) = \lim_{x \to -3} \frac{f(x) - f(-3)}{x + 3}
\]

### Step 1: Compute \( f(-3) \)

First, we need to evaluate \( f(-3) \):

\[
f(-3) = \frac{1}{-3 - 3} = \frac{1}{-6} = -\frac{1}{6}
\]

### Step 2: Set up the limit

Now substitute \( f(x) = \frac{1}{x - 3} \) and \( f(-3) = -\frac{1}{6} \) into the limit definition of the derivative:

\[
f'(-3) = \lim_{x \to -3} \frac{\frac{1}{x - 3} - \left( -\frac{1}{6} \right)}{x + 3}
\]

This simplifies to:

\[
f'(-3) = \lim_{x \to -3} \frac{\frac{1}{x - 3} + \frac{1}{6}}{x + 3}
\]

### Step 3: Combine the fractions

To combine the fractions in the numerator, we need a common denominator. The common denominator is \( 6(x - 3) \), so we rewrite the numerator as:

\[
\frac{1}{x - 3} + \frac{1}{6} = \frac{6 + (x - 3)}{6(x - 3)} = \frac{x + 3}{6(x - 3)}
\]

### Step 4: Simplify the expression

Now substitute this back into the limit:

\[
f'(-3) = \lim_{x \to -3} \frac{\frac{x + 3}{6(x - 3)}}{x + 3}
\]

This simplifies to:

\[
f'(-3) = \lim_{x \to -3} \frac{1}{6(x - 3)}
\]

### Step 5: Evaluate the limit

As \( x \to -3 \), \( x - 3 \to -6 \). So we have:

\[
f'(-3) = \frac{1}{6(-6)} = \frac{1}{-36} = -\frac{1}{36}
\]

### Final Answer:
\[
f'(-3) = -\frac{1}{36}
\]

Would you like further details on any step, or have any questions?

Here are 5 related questions to deepen understanding:
1. How is the derivative related to the slope of the tangent line?
2. Why do we use limits to find the derivative?
3. How does the derivative of \( f(x) = \frac{1}{x-3} \) change as you vary the point \( x = a \)?
4. What does the negative slope tell you about the function at \( x = -3 \)?
5. Can you apply the same method to find the derivative of \( f(x) = \frac{1}{x+2} \)?

**Tip**: When combining fractions, always look for a common denominator to simplify the expression effectively!

The slope of the tangent line to f(x) at x=a can be found by evaluating the limit f'(a)=mtan=limx->a f(x)-f(a)/x-a provided the limit exists. Use this limit to find f'(-3) given f(x)=1/x-3. Find f'(-3).

Learn how to find the derivative f'(-3) for the function f(x) = 1/(x - 3) using the limit definition of a derivative. This step-by-step solution covers key calculus concepts such as limits, tangent lines, and derivative computation. Suitable for students in Grades 11-12.

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