Question

1. the function, its 1st and 2nd derivatives are given

$f(x) = \frac{4x}{x^2 + 1}$, \quad $f(x) = \frac{4(1 - x^2)}{(x^2 + 1)^2}$, \quad $f(x) = \frac{8x(x^2 - 3)}{(x^2 + 1)^3}$,
find the intercepts, asymptotes, local max & mins, the points of inflection,
and use this information to sketch the graph of $f$.

Explanation:

Step1: Find Intercepts

• x-intercept: Set \( f(x) = 0 \), so \( \frac{4x}{x^2 + 1}=0 \). Since \( x^2 + 1

eq0 \) for all real \( x \), solve \( 4x = 0 \), giving \( x = 0 \). So x-intercept is \( (0, 0) \).

• y-intercept: Set \( x = 0 \) in \( f(x) \), \( f(0)=\frac{0}{0 + 1}=0 \). So y-intercept is \( (0, 0) \).

Step2: Find Asymptotes

• Vertical Asymptotes: Denominator \( x^2 + 1 \) has no real roots (since \( x^2=-1 \) has no real solutions), so no vertical asymptotes.
• Horizontal Asymptotes: Compute \( \lim_{x\to\pm\infty}f(x)=\lim_{x\to\pm\infty}\frac{4x}{x^2 + 1}=\lim_{x\to\pm\infty}\frac{4/x}{1 + 1/x^2}=0 \). So horizontal asymptote is \( y = 0 \).

Step3: Find Local Max/Min (Using \( f'(x) \))

• Set \( f'(x)=0 \): \( \frac{4(1 - x^2)}{(x^2 + 1)^2}=0 \). Numerator \( 4(1 - x^2)=0\Rightarrow1 - x^2=0\Rightarrow x=\pm1 \).
• Test intervals:
• For \( x < -1 \), say \( x=-2 \), \( f'(-2)=\frac{4(1 - 4)}{(4 + 1)^2}=\frac{-12}{25}<0 \) (decreasing).
• For \( -1 < x < 1 \), say \( x = 0 \), \( f'(0)=\frac{4(1 - 0)}{(0 + 1)^2}=4>0 \) (increasing).
• For \( x > 1 \), say \( x = 2 \), \( f'(2)=\frac{4(1 - 4)}{(4 + 1)^2}=\frac{-12}{25}<0 \) (decreasing).
• At \( x=-1 \): \( f(-1)=\frac{-4}{1 + 1}=-2 \) (local min, since function changes from decreasing to increasing).
• At \( x = 1 \): \( f(1)=\frac{4}{1 + 1}=2 \) (local max, since function changes from increasing to decreasing).

Step4: Find Points of Inflection (Using \( f''(x) \))

• Set \( f''(x)=0 \): \( \frac{8x(x^2 - 3)}{(x^2 + 1)^3}=0 \). Numerator \( 8x(x^2 - 3)=0\Rightarrow x = 0 \) or \( x=\pm\sqrt{3} \).
• Test concavity:
• For \( x < -\sqrt{3} \), say \( x=-2 \), \( f''(-2)=\frac{8(-2)(4 - 3)}{(4 + 1)^3}=\frac{-16}{125}<0 \) (concave down).
• For \( -\sqrt{3}
• For \( 0 < x < \sqrt{3} \), say \( x = 1 \), \( f''(1)=\frac{8(1)(1 - 3)}{(1 + 1)^3}=\frac{-16}{8}=-2<0 \) (concave down).
• For \( x > \sqrt{3} \), say \( x = 2 \), \( f''(2)=\frac{8(2)(4 - 3)}{(4 + 1)^3}=\frac{16}{125}>0 \) (concave up).
• Find \( f(x) \) at these points:
• \( f(0)=0 \), \( f(\sqrt{3})=\frac{4\sqrt{3}}{3 + 1}=\sqrt{3} \), \( f(-\sqrt{3})=\frac{-4\sqrt{3}}{3 + 1}=-\sqrt{3} \).
• So inflection points: \( (-\sqrt{3}, -\sqrt{3}) \), \( (0, 0) \), \( (\sqrt{3}, \sqrt{3}) \).

Step5: Sketch the Graph

• Plot intercept \( (0,0) \), local min \( (-1, -2) \), local max \( (1, 2) \), inflection points \( (-\sqrt{3}, -\sqrt{3}) \), \( (0, 0) \), \( (\sqrt{3}, \sqrt{3}) \).
• Draw horizontal asymptote \( y = 0 \).
• The graph increases from \( -\infty \) to \( x = 1 \) (with a local min at \( x=-1 \)), then decreases to \( \infty \), with concavity changing at \( x=\pm\sqrt{3} \) and \( x = 0 \).

Answer:

• Intercepts: \( (0, 0) \) (both x and y-intercept).
• Asymptotes: Horizontal asymptote \( y = 0 \), no vertical asymptotes.
• Local Max/Min: Local max at \( (1, 2) \), local min at \( (-1, -2) \).
• Inflection Points: \( (-\sqrt{3}, -\sqrt{3}) \), \( (0, 0) \), \( (\sqrt{3}, \sqrt{3}) \).
• Graph Sketch: Plot the above points, draw horizontal asymptote \( y = 0 \), show increasing/decreasing and concavity changes as analyzed.