Question
Consider the family of functions \(f(x)=\frac{1}{x^{2}-2x+k},\) where k is a constant.
(a) Find the value of k, for k > 0, such that the slope of the line tangent to the graph of f at x = 0 equals 6.
(b) For k = −8, find the value of \(\int_{0}^{1}f(x)dx.\)
(c) For k = 1, find the value of \(\int_{0}^{1}f(x)dx\) or show that it diverges.
Answer/Explanation
Ans:
(a) \(f'(x)=\frac{-(2x-2)}{\left ( x^{2}-2x+k \right )^{2}}\)
\(f'(0)=\frac{2}{k^{2}}=6\Rightarrow k^{2}=\frac{1}{3}\Rightarrow k = \frac{1}{\sqrt{3}}\)
(b) \(\frac{1}{x^{2}-2x-8}=\frac{1}{(x-4)(x+2)}=\frac{A}{x-4}+\frac{B}{x+2}\)
\(\Rightarrow 1 = A(x+2)+B(x-4)\)
\(\Rightarrow A = \frac{1}{6}, B=-\frac{1}{6}\)
\(\int_{0}^{1}f(x)dx=\int_{0}^{1}\left ( \frac{\frac{1}{6}}{x-4} -\frac{\frac{1}{6}}{x+2}\right )dx\)
\(=\left [\frac{1}{6}In|x-4| -\frac{1}{6} In|x+2| \right ]_{x=0}^{x=1}\)
\(=\left ( \frac{1}{6}In3-\frac{1}{6}In3 \right )-\left ( \frac{1}{6}In4-\frac{1}{6}In2 \right )=-\frac{1}{6}In2\)
(c) \(\int_{0}^{2}\frac{1}{x^{2}-2x+1}dx=\int_{0}^{2}\frac{1}{(x-1)^{2}}dx=\int_{0}^{1}\frac{1}{(x-1)^{2}}dx+\int_{1}^{2}\frac{1}{(x-1)^{2}}dx\)
\(=\lim_{b\rightarrow 1^{-}}\int_{0}^{b}\frac{1}{(x-1)^{2}}dx+\lim_{b\rightarrow 1^{+}}\int_{b}^{2}\frac{1}{(x-1)^{2}}dx\)
\(=\lim_{b\rightarrow 1^{-}}\left ( -\frac{1}{x-1|_{x=0}^{x=b}} \right )+\lim_{b\rightarrow 1^{+}}\left ( -\frac{1}{x-1|_{x=b}^{x=2}} \right )\)
\(=\lim_{b\rightarrow 1^{-}}\left ( -\frac{1}{b-1}-1 \right )+\lim_{b\rightarrow 1^{+}}\left ( -1+\frac{1}{b-1} \right )\)
Because \(=\lim_{b\rightarrow 1^{-}}\left ( -\frac{1}{b-1} \right )\) does not exist, the integral diverges.
Question
Consider the function \(f(x)=\frac{1}{x^{2}-kx},\) where k is a nonzero constant. The derivative of f is given by \(f'(x)=\frac{k-2x}{\left ( x^{2}-kx \right )^{2}}.\)
(a) Let k = 3, so that \(f(x)=\frac{1}{x^{2}-3x}.\) Write an equation for the line tangent to the graph of f at the point whose x-coordinate is 4.
(b) Let k = 4, so that \(f(x)=\frac{1}{x^{2}-4x}.\) Determine whether f has a relative minimum, a relative maximum, or neither at x = 2. Justify your answer.
(c) Find the value of k for which f has a critical point at x = -5.
(d) Let k = 6, so that \(f(x)=\frac{1}{x^{2}-6x}\) Find the partial fraction decomposition for the function f. Find \(\int f(x)dx.\)
Answer/Explanation
Ans:
(a) \(f(4)=\frac{1}{4^{2}-3\cdot 4}=\frac{1}{4}\) \(f'(4)=\frac{3-2\cdot 4}{\left ( 4^{2}-3\cdot 4 \right )^{2}}=-\frac{5}{16}\)
An equation for the line tangent to the graph of f at the point whose x-coordinate is 4 is \(y= -\frac{5}{16}(x-4)+\frac{1}{4}.\)
(b) \(f'(x)=\frac{4-2x}{\left ( x^{2}-4x \right )^{2}}\) \(f'(2)=\frac{4-2\cdot 2}{\left ( 2^{2}-4\cdot 2 \right )^{2}}=0\)
f'(x) changes sign from positive to negative at x = 2. Therefore, f has a relative maximum at x = 2.
(c) \(f'(-5)=\frac{k-2\cdot (5)}{\left ( (-5)^{2}-k\cdot (-5) \right )^{2}}=0\Rightarrow k = -10\)
(d) \(\frac{1}{x^{2}-6x}=\frac{1}{x(x-6)}=\frac{A}{x}+\frac{B}{x-6}\Rightarrow 1=A(x-6)+Bx\)
\(x = 0 \Rightarrow 1 = A\cdot (-6)\Rightarrow A = -\frac{1}{6}\)
\(x = 6 \Rightarrow 1 = B\cdot (6)\Rightarrow B = \frac{1}{6}\)
\(\frac{1}{x(c-6)}=\frac{-1/6}{x}+\frac{1/6}{x-6}\)
\(\int f(x)dx-\int \left ( \frac{-1/6}{x}+\frac{1/6}{x-6} \right )dx\)
\(=-\frac{1}{6}In|x| +\frac{1}{6}In|x-6|+C=\frac{1}{6}In\left | \frac{x-6}{x} \right |+C\)