Use the alternate form of the derivative to find the derivative at x = c, if it exists.

 f(x) = x^3 + 2x^2 + 1, c = -2

The Correct Answer and Explanation is:

We are asked to find the derivative of the function
f(x) = x³ + 2x² + 1
at the point x = c = -2 using the alternate form of the derivative.

Step 1: Use the Alternate Form of the Derivative

The alternate form of the derivative is:f′(c)=lim⁡x→cf(x)−f(c)x−cf'(c) = \lim_{x \to c} \frac{f(x) – f(c)}{x – c}f′(c)=x→clim​x−cf(x)−f(c)​

Here, f(x)=x3+2×2+1f(x) = x^3 + 2x^2 + 1f(x)=x3+2×2+1 and c=−2c = -2c=−2.

Step 2: Compute f(c)f(c)f(c)

f(−2)=(−2)3+2(−2)2+1=−8+8+1=1f(-2) = (-2)^3 + 2(-2)^2 + 1 = -8 + 8 + 1 = 1f(−2)=(−2)3+2(−2)2+1=−8+8+1=1

Step 3: Plug into the Formula

f′(−2)=lim⁡x→−2f(x)−f(−2)x+2=lim⁡x→−2×3+2×2+1−1x+2=lim⁡x→−2×3+2x2x+2f'(-2) = \lim_{x \to -2} \frac{f(x) – f(-2)}{x + 2} = \lim_{x \to -2} \frac{x^3 + 2x^2 + 1 – 1}{x + 2} = \lim_{x \to -2} \frac{x^3 + 2x^2}{x + 2}f′(−2)=x→−2lim​x+2f(x)−f(−2)​=x→−2lim​x+2×3+2×2+1−1​=x→−2lim​x+2×3+2×2​

Step 4: Simplify the Expression

Factor the numerator:x3+2×2=x2(x+2)x^3 + 2x^2 = x^2(x + 2)x3+2×2=x2(x+2)

So the limit becomes:f′(−2)=lim⁡x→−2×2(x+2)x+2f'(-2) = \lim_{x \to -2} \frac{x^2(x + 2)}{x + 2}f′(−2)=x→−2lim​x+2×2(x+2)​

Cancel x+2x + 2x+2 (as long as x≠−2x \ne -2x=−2):=lim⁡x→−2×2= \lim_{x \to -2} x^2=x→−2lim​x2=(−2)2=4= (-2)^2 = 4=(−2)2=4

Final Answer:

f′(−2)=4\boxed{f'(-2) = 4}f′(−2)=4​

Explanation

To find the derivative of a function at a point using the alternate form, we apply the definition:f′(c)=lim⁡x→cf(x)−f(c)x−cf'(c) = \lim_{x \to c} \frac{f(x) – f(c)}{x – c}f′(c)=x→clim​x−cf(x)−f(c)​

This expression gives the slope of the tangent line to the curve at the point x=cx = cx=c. Instead of using rules like the power rule, this method uses limits to find the derivative from the basic definition.

In our example, the function is f(x)=x3+2×2+1f(x) = x^3 + 2x^2 + 1f(x)=x3+2×2+1, and we are finding the derivative at c=−2c = -2c=−2. First, we evaluate f(−2)f(-2)f(−2) and substitute into the formula. Simplifying the expression gives us a rational function with a removable discontinuity at x=−2x = -2x=−2. By factoring the numerator, we can cancel out the x+2x + 2x+2 term in both numerator and denominator, which resolves the indeterminate form.

After canceling, we evaluate the remaining expression x2x^2×2 at x=−2x = -2x=−2, which results in 444. This is the slope of the tangent line to the curve at the point x=−2x = -2x=−2.

This method is valuable for understanding what a derivative truly means: the instantaneous rate of change of a function at a specific point. It also reinforces the concept of limits and how algebraic simplification can help us evaluate them when direct substitution gives an indeterminate form.