Problem 74 - Calculus - AP Edition

Question

$$ \begin{array}{l} \text { Let } f(x)=x^{8}-2 x+3 ; \text { find } \\ \qquad \lim _{w \rightarrow 2} \frac{f^{\prime}(w)-f^{\prime}(2)}{w-2} \end{array} $$

Step-by-Step Solution

Answer

The limit is 3584.

Step 1: Understand the Problem

We are tasked with finding the limit of a given expression as \( w \) approaches 2. The expression involves the derivative of the function \( f(x) = x^8 - 2x + 3 \). We need to analyze the behavior of \( \frac{f'(w) - f'(2)}{w-2} \) as \( w \) gets closer to 2.

Step 2: Differentiate the Function

Calculate the derivative of \( f(x) = x^8 - 2x + 3 \). Applying the power rule and constant rule, we find:\( f'(x) = 8x^7 - 2 \).

Step 3: Evaluate the Derivative at w = 2

Substitute \( x = 2 \) in the derivative \( f'(x) = 8x^7 - 2 \):\( f'(2) = 8(2)^7 - 2 = 8 \times 128 - 2 = 1024 - 2 = 1022 \).

Step 4: Substitute Derivative into Limit Expression

The limit expression is:\( \lim_{w \to 2} \frac{f'(w) - f'(2)}{w-2} = \lim_{w \to 2} \frac{8w^7 - 2 - 1022}{w-2} \).

Step 5: Simplify the Numerator

The numerator simplifies to \( 8w^7 - 1024 \). We need to factor this expression to simplify the limit. Notice that \( 8w^7 - 1024 = 8(w^7 - 128) \).

Step 6: Factor the Expression

The expression \( w^7 - 128 \) can be factored using the polynomial identity for the difference of powers. Specifically, \( w^7 - 128 = (w-2)(w^6 + 2w^5 + 4w^4 + 8w^3 + 16w^2 + 32w + 64) \).

Step 7: Simplify the Limit Expression

Substitute the factored form into the limit expression:\( \lim_{w \to 2} \frac{8(w-2)(w^6 + 2w^5 + 4w^4 + 8w^3 + 16w^2 + 32w + 64)}{w-2} \).Cancel the \( (w-2) \) terms:\( \lim_{w \to 2} 8(w^6 + 2w^5 + 4w^4 + 8w^3 + 16w^2 + 32w + 64) \).

Step 8: Evaluate the Limit

Now that we have canceled the common factor, evaluate the polynomial at \( w = 2 \):\( 8((2)^6 + 2(2)^5 + 4(2)^4 + 8(2)^3 + 16(2)^2 + 32(2) + 64) \).Calculate:- \( 2^6 = 64 \)- \( 2^5 = 32 \)- \( 2^4 = 16 \)- \( 2^3 = 8 \)- \( 2^2 = 4 \)This gives us:\( 8(64 + 2 \cdot 32 + 4 \cdot 16 + 8 \cdot 8 + 16 \cdot 4 + 32 \cdot 2 + 64) \).Calculate each:- \( 2 \cdot 32 = 64 \)- \( 4 \cdot 16 = 64 \)- \( 8 \cdot 8 = 64 \)- \( 16 \cdot 4 = 64 \)- \( 32 \cdot 2 = 64 \)Add all these to find:\( 8(64 + 64 + 64 + 64 + 64 + 64 + 64) = 8 \cdot 448 = 3584 \).

Step 9: Conclusion

The desired limit evaluates to 3584.

Key Concepts

Polynomial Derivatives

In calculus, one often encounters polynomial functions, like the given function \( f(x) = x^8 - 2x + 3 \), and finds their derivatives to analyze rate changes, slopes of tangent lines, and other dynamic behavior of functions. Derivatives of polynomial functions, like \( f'(x) \), are calculated by applying differentiation rules to the polynomial's individual terms.

The derivative of a polynomial is the sum of the derivatives of its terms. For instance, to find the derivative of \( f(x) \), you take the derivative term-by-term: the derivative of \( x^8 \), the derivative of \( -2x \), and the derivative of the constant \( 3 \). Remember that the derivative of a constant, like \( 3 \), is zero because constants do not change. Hence, applying calculus rules, \( f'(x) = 8x^7 - 2 \).

Derivatives provide essential data about a function's instantaneous rate of change, which plays a crucial role in many areas like physics, engineering, and economics.

Power Rule in Calculus

The power rule is a straightforward and essential tool in calculus, making it easy to find the derivative of polynomial terms. It is expressed mathematically as: if you have a function \( f(x) = x^n \), where \( n \) is a constant, then the derivative \( f'(x) \) is \( nx^{n-1} \).

This rule is applied term-by-term to each power of \( x \) in a polynomial. For our original function \( f(x) = x^8 - 2x + 3 \):

• Applying the power rule to \( x^8 \) gives \( 8x^7 \).
• The term \( -2x \) simplifies using the power rule to \( -2 \times 1x^{1-1} = -2 \).
• The constant \( 3 \), as mentioned earlier, becomes \( 0 \).

This results in the derivative \( f'(x) = 8x^7 - 2 \), showing how the power rule simplifies the differentiation process for polynomial functions. Luckily, this rule can be applied quickly even in much more complicated polynomial expressions.

Factoring Polynomials

Factoring polynomials involves breaking down a polynomial into simpler parts, often into products of lower-degree polynomials. This skill is invaluable in solving equations, simplifying expressions, and analyzing polynomial behavior.

In our solution, it was necessary to factor the expression \( w^7 - 128 \) to simplify the limit. This expression can be seen as a difference of powers, specifically \( w^7 \) and \( 128 = 2^7 \). You can use the identity for factoring the difference of powers, which allows us to express this as:

• \( (w-2) \) is one factor.
• The other factor is a polynomial of one less degree, \( w^6 + 2w^5 + 4w^4 + 8w^3 + 16w^2 + 32w + 64 \).

This factorization enables us to cancel the \( (w-2) \) terms in the limit expression, simplifying the calculation. Mastering polynomial factorization helps in not just calculus but also algebra, by solving and understanding polynomial equations better.