Lesson Couple of non-standard Calculus problems
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<H2>Couple of non-standard Calculus problems</H2> <H3>Problem 1</H3>If {{{(f(x))^(2n)}}} + {{{((f(x) - 2))^(2n)}}} = {{{x^(2n)}}}, where n is a natural number, then find (dx)/(d(f(x))) at f(x) = 1. <B>Solution</B> <pre> So, n is a natural number, i.e. positive integer number. Differentiate the given equation. You will get (2n)*f^(2n-1)(x)*f'(x) + (2n)*(f(x)-2)^(2n-1)*f'(x) = (2n)*x^(2n-1). Cancel factor (2n) in both sides f^(2n-1)(x)*f'(x) + (f(x)-2)^(2n-1)*f'(x) = x^(2n-1). Take f'(x) out the parentheses as a common factor df ---- * [f^(2n-1)(x) + (f(x)-2)^(2n-1)] = x^(2n-1). dx Then dx f^(2n-1)(x) + (f(x)-2)^(2n-1)) ---- = ---------------------------------- df x^(2n-1) Take it at f(x) = 1 dx 1^(2n-1) + (1-2)^(2n-1) 1 - 1 ---- = -------------------------- = ----------. df x^(2n-1) x^(2n-1) dx So, if x =/= 0, then ---- = 0. df But x definitely is not zero, since, otherwise, in the original equation left side would be {{{1^(2n) + (-1)^(2n))}}} = 1 + 1 = 2, while the right side would be {{{0^(2n)}}} = 0. dx Thus, the final conclusion is that ---- = 0. <U>ANSWER</U> df </pre> <H3>Problem 2</H3>If f(x) = {{{(k+x^3sec(x))/sin(b*(x-pi))}}} is continuous at {{{x=pi}}}, and {{{kb - pi*f(pi)}}} = {{{4*pi^3}}}, find the value of b. <B>Solution</B> <pre> sec(x) = {{{1/cos(x)}}}. When x approaches to π, then cos(x) approaches to -1; hence, sec(x) approaches to -1. It means that k+x^3*sec(x) in the numerator is close to k-x^3. In the denominator, sin(b) is a non-zero constant, so, we should not worry about it. But in the denominator, we have x-π, and, in order for the expression be a continuous function, k-x^3 in the numerator must put out this x-π in the denominator. Hence, k must be π^3. Then the numerator, as x approaches to π, is equivalent to π^3-x^3 = (π-x)*(π^2 +πx + x^2), and it puts out x-π in the denominator. So, we deduced logically that the expression is a continuous function if and only if k = π^3. Then kb = {{{pi^3*b}}}, f(π) = - {{{((pi^2 + pi^2 + pi^2))/b}}} = - {{{(3*pi^2)/b}}}; and kb - π*f(π) = {{{pi^3*b}}} + {{{(3*pi^3)/b}}}. It gives us this equation {{{pi^3*b}}} + {{{(3*pi^3)/b}}} = {{{4*pi^3}}}. and we want to solve it for "b". Divide all the terms by {{{pi^3}}} to get {{{b}}} + {{{3/b}}} = 4. Multiply both sides by b. You will get this quadratic equation b^2 - 4b + 3 = 0. Factor left side (b-3)*(b-1) = 0. <U>ANSWER</U>. 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