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
Let f be the function defined by \(f(x)=\frac{3}{2x^{2}-7x+5}.\)
(a) Find the slope of the line tangent to the graph of f at x = 3.
(b) Find the x-coordinate of each critical point of f in the interval 1 < x < 2.5. Classify each critical point as the location of a relative minimum, a relative maximum, or neither. Justify your answers.
(c) Using the identity that \(\frac{3}{2x^{2}-7x+5}=\frac{2}{2x-5}-\frac{1}{x-1}, evaluate\int_{5}^{\infty }f(x)\) or show that the integral diverges.
(d) Determine whether the series \(\sum_{n=5}^{\infty }\frac{3}{2n^{2}-7n+5}\) converges or diverges. State the conditions of the test used for determining convergence or divergence.
Answer/Explanation
Ans:
(a) \(f'(x)=\frac{-3(4x-7)}{\left ( 2x^{2}-7x+5 \right )^{2}}\)
\(f'(3)=\frac{-3(5)}{\left ( 18-21+5 \right )^{2}}=-\frac{15}{4}\)
(b) \(f'(x)=\frac{-3(4x-7)}{\left ( 2x^{2}-7x+5 \right )^{2}}=0\Rightarrow x=\frac{7}{4}\)
The only critical point in the interval 1 < x < 2.5 has x-coordinate \(\frac{7}{4}.\) f ′ changes sign from positive to negative at \(x=\frac{7}{4}.\) Therefore, f has a relative maximum at \(x=\frac{7}{4}.\)
(c) \(\int_{5}^{\infty }f(x)dx=\lim_{b\rightarrow \infty }\int_{5}^{b}\frac{3}{2x^{2}-7x+5}dx =\lim_{b\rightarrow \infty }\int_{5}^{b}\left ( \frac{2}{2x-5} -\frac{1}{x-1}\right )dx\)
\(=\lim_{b\rightarrow \infty }\left [ In(2x-5)-In(x-1) \right ]_{5}^{b}=\lim_{b\rightarrow \infty }\left [ In\left ( \frac{2x-5}{x-1} \right ) \right ]_{5}^{b}\)
\(=\lim_{b\rightarrow \infty }\left [ In(\frac{2b-5}{b-1})-In(\frac{5}{4}) \right ]=In 2-In\left ( \frac{5}{4} \right )=In\left ( \frac{8}{5} \right )\)
(d) f is continuous, positive, and decreasing on [5, ∞). The series converges by the integral test since \(\int_{5}^{\infty }\frac{3}{2x^{2}-7x+5}dx\) converges.
-OR –
\(\frac{3}{2n^{2}-7n+5}>0 and \frac{1}{n^{2}}>0 for n\geq 5.\)
Since \(\frac{\frac{3}{2n^{2}-7n+5}}{\frac{1}{n^{2}}} =\frac{3}{2}\) and the series \(\sum_{n=5}^{\infty }\frac{1}{n^{2}}\) converges, the series \(\sum_{n=5}^{\infty }\frac{3}{2n^{2}-7n+5}\) converges by the limit comparison test.