Question
Using the definition, calculate the derivative of the function. Then find the values of the derivative as specified. \( f(x)=2+x^{2} ; f^{\prime}(-3), f^{\prime}(0), f^{\prime}(1) \)
Ask by Sanders Lane. in the United States
Mar 09,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The derivative of \( f(x) = 2 + x^2 \) is \( f'(x) = 2x \). Evaluating at the specified points:
- \( f'(-3) = -6 \)
- \( f'(0) = 0 \)
- \( f'(1) = 2 \)
Solution
**Step 1: Use the Definition of the Derivative**
The definition of the derivative is
\[
f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}.
\]
**Step 2: Write \( f(x+h) \) for \( f(x)=2+x^2 \)**
Substitute \( x+h \) into the function:
\[
f(x+h)=2+(x+h)^2 = 2 + x^2 + 2xh + h^2.
\]
**Step 3: Compute \( f(x+h)-f(x) \)**
Subtract \( f(x) \) from \( f(x+h) \):
\[
f(x+h)-f(x) = \left[ 2+x^2+2xh+h^2 \right] - \left[ 2+x^2 \right] = 2xh+h^2.
\]
**Step 4: Form the Difference Quotient**
Divide by \( h \):
\[
\frac{f(x+h)-f(x)}{h} = \frac{2xh+h^2}{h} = 2x + h.
\]
**Step 5: Take the Limit as \( h \to 0 \)**
\[
f'(x)=\lim_{h \to 0}(2x+h)=2x.
\]
Thus, the derivative of the function is
\[
f'(x)=2x.
\]
**Step 6: Evaluate the Derivative at the Specified Points**
1. For \( x=-3 \):
\[
f'(-3) = 2(-3) = -6.
\]
2. For \( x=0 \):
\[
f'(0) = 2(0) = 0.
\]
3. For \( x=1 \):
\[
f'(1) = 2(1) = 2.
\]
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Simplify this solution Bonus Knowledge
The derivative \( f'(x) \) of the function \( f(x) = 2 + x^{2} \) can be calculated using the power rule. The derivative of \( x^2 \) is \( 2x \), so we have: \[ f'(x) = 0 + 2x = 2x \] Now, let's evaluate the derivative at the specified values: 1. \( f'(-3) = 2(-3) = -6 \) 2. \( f'(0) = 2(0) = 0 \) 3. \( f'(1) = 2(1) = 2 \) So, the values of the derivative are: - \( f'(-3) = -6 \) - \( f'(0) = 0 \) - \( f'(1) = 2 \)
