Math Problem Statement

Consider functions 𝑓(π‘₯) = βˆ‘ π‘₯ 𝑛 𝑖 𝑖=0 (= (π‘₯ 𝑛+1βˆ’1) π‘₯βˆ’1 , for π‘₯ β‰  1) and 𝑔(π‘₯) = (π‘₯ βˆ’ 1) 2 . (a) Compute the first derivative 𝑓′(π‘₯) of function 𝑓(π‘₯) represented as a summation. (b) What is the value of 𝑓 β€² (π‘₯) for π‘₯ = 1 and π‘₯ β‰  1?

Solution

Given the functions:

f(x)=βˆ‘i=0nxi=xn+1βˆ’1xβˆ’1for xβ‰ 1f(x) = \sum_{i=0}^{n} x^i = \frac{x^{n+1} - 1}{x - 1} \quad \text{for } x \neq 1 and g(x)=(xβˆ’1)2g(x) = (x - 1)^2

(a) Compute the First Derivative fβ€²(x)f'(x) of f(x)f(x)

Since f(x)=xn+1βˆ’1xβˆ’1f(x) = \frac{x^{n+1} - 1}{x - 1}, we can find fβ€²(x)f'(x) using the quotient rule for derivatives. For h(x)=xn+1βˆ’1h(x) = x^{n+1} - 1 and k(x)=xβˆ’1k(x) = x - 1:

  1. h(x)=xn+1βˆ’1β‡’hβ€²(x)=(n+1)xnh(x) = x^{n+1} - 1 \Rightarrow h'(x) = (n+1)x^n
  2. k(x)=xβˆ’1β‡’kβ€²(x)=1k(x) = x - 1 \Rightarrow k'(x) = 1

Using the quotient rule:

fβ€²(x)=hβ€²(x)k(x)βˆ’h(x)kβ€²(x)(k(x))2=(n+1)xn(xβˆ’1)βˆ’(xn+1βˆ’1)(xβˆ’1)2f'(x) = \frac{h'(x)k(x) - h(x)k'(x)}{(k(x))^2} = \frac{(n+1)x^n (x - 1) - (x^{n+1} - 1)}{(x - 1)^2}

Expanding and simplifying:

fβ€²(x)=(n+1)xn+1βˆ’(n+1)xnβˆ’xn+1+1(xβˆ’1)2f'(x) = \frac{(n+1)x^{n+1} - (n+1)x^n - x^{n+1} + 1}{(x - 1)^2} =nxn+1βˆ’(n+1)xn+1(xβˆ’1)2= \frac{n x^{n+1} - (n+1)x^n + 1}{(x - 1)^2}

(b) Compute f′(x)f'(x) for x=1x = 1 and x≠1x \neq 1

For x≠1x \neq 1:

Using the simplified expression obtained above, we have:

fβ€²(x)=nxn+1βˆ’(n+1)xn+1(xβˆ’1)2for xβ‰ 1f'(x) = \frac{n x^{n+1} - (n+1)x^n + 1}{(x - 1)^2} \quad \text{for } x \neq 1

For x=1x = 1:

To find fβ€²(1)f'(1), we use L'HΓ΄pital's rule, since the expression f(x)=xn+1βˆ’1xβˆ’1f(x) = \frac{x^{n+1} - 1}{x - 1} is indeterminate at x=1x = 1.

  1. Differentiate the numerator and denominator:
    • The derivative of xn+1βˆ’1x^{n+1} - 1 is (n+1)xn(n+1)x^n.
    • The derivative of xβˆ’1x - 1 is 11.

Then: fβ€²(1)=lim⁑xβ†’1(n+1)xn1=n+1f'(1) = \lim_{x \to 1} \frac{(n+1)x^n}{1} = n + 1

So:

  • For xβ‰ 1x \neq 1: fβ€²(x)=nxn+1βˆ’(n+1)xn+1(xβˆ’1)2f'(x) = \frac{n x^{n+1} - (n+1)x^n + 1}{(x - 1)^2}
  • For x=1x = 1: fβ€²(1)=n+1f'(1) = n + 1

Would you like more details on this solution or have further questions?

Here are some follow-up questions you might consider:

  1. How is the quotient rule applied in general, and why is it useful here?
  2. What are the conditions for using L'HΓ΄pital's rule in calculating limits?
  3. How does the function g(x)=(xβˆ’1)2g(x) = (x-1)^2 relate to the behavior of f(x)f(x) as xx approaches 1?
  4. Can the derivative fβ€²(x)f'(x) be interpreted graphically around x=1x = 1?
  5. What happens to f(x)f(x) as nβ†’βˆžn \to \infty for various values of xx?

Tip: When dealing with indeterminate forms, L'HΓ΄pital's rule is often a reliable tool to simplify limits.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Summation
L'HΓ΄pital's Rule
Quotient Rule

Formulas

f(x) = (x^(n+1) - 1) / (x - 1)
g(x) = (x - 1)^2
Quotient Rule: (h(x)/k(x))' = (h'(x)k(x) - h(x)k'(x)) / (k(x))^2
L'HΓ΄pital's Rule: lim(x β†’ c) f(x)/g(x) = lim(x β†’ c) f'(x)/g'(x) when lim(x β†’ c) f(x)/g(x) is indeterminate

Theorems

Quotient Rule
L'HΓ΄pital's Rule

Suitable Grade Level

College Level - Calculus I