If f(x)=`x^2` , then `(f(x+h)-f(x))/(h)`=

Options

A. 0
B. h
C. 2x
D. 2x+h

Explanation

To solve the problem, we need to evaluate the expression \((f(x+h) - f(x)) / h\) given that \(f(x) = x^2\).

### Step-by-step Solution:

1. **Substitute \(f(x)\) and \(f(x+h)\)**:
- We know \(f(x) = x^2\).
- Now, we need to find \(f(x+h)\):
\[ f(x+h) = (x+h)^2 \]

2. **Expand \(f(x+h)\)**:
- Using the formula for the square of a binomial, \((a+b)^2 = a^2 + 2ab + b^2\):
\[ f(x+h) = (x+h)^2 = x^2 + 2xh + h^2 \]

3. **Set up the expression**:
- Now we substitute \(f(x+h)\) and \(f(x)\) into the expression:
\[ \frac{f(x+h) - f(x)}{h} = \frac{(x^2 + 2xh + h^2) - x^2}{h} \]

4. **Simplify the numerator**:
- The \(x^2\) terms cancel out:
\[ = \frac{2xh + h^2}{h} \]

5. **Factor out \(h\) from the numerator**:
- We can factor \(h\) out of the numerator:
\[ = \frac{h(2x + h)}{h} \]

6. **Cancel \(h\) in the numerator and denominator**:
- As long as \(h \neq 0\), we can cancel \(h\):
\[ = 2x + h \]

### Final Answer:
Thus, the expression \(\frac{f(x+h) - f(x)}{h}\) simplifies to:
\[ 2x + h \]