Area Under the Curve
Area functional equation
Grade 12
Question:
<p>Suppose \(f:[-1,1]\to\mathbb{R}\) is defined by \(f(x)=\dfrac{x^2}{x+1}\) where \(f\) is defined at \(x\ne-1\). Which statements hold? [MAU037]</p>
<li>\(f\) has exactly one zero in \((0,1)\)</li>
<li>\(f\) is many-one on \([-1,1]\)</li>
<li>\(f\) is increasing for \(x>0\) and decreasing for \(x<0\)</li>
<li>\(f\) has a minimum not attained on \([-1,1]\)</li>
Step-by-Step Solution
Key Concept: f(x)=x^2/(x+1). Zero at x=0. f'(x)=(x(x+2))/(x+1)^2. For x\in (0,1): f'>0 (increasing). The area under f from 0 to 1 is straightforward.
<div class='solution'>
<p>$f(x)=\frac{x^2}{x+1}=x-1+\frac{1}{x+1}$.</p>
<p>$f'(x)=1-\frac{1}{(x+1)^2}=\frac{(x+1)^2-1}{(x+1)^2}=\frac{x^2+2x}{(x+1)^2}=\frac{x(x+2)}{(x+1)^2}$.</p>
<p><strong>A:</strong> $f(0)=0$, $f(1)=1/2>0$, $f$ has exactly one zero at $x=0$. In $(0,1)$: zero only at boundary. ✗ (zero at x=0 is endpoint). Interpret as: only zero on $[-1,1]$ is $x=0$. A depends on exact question wording.</p>
<p><strong>D:</strong> Near $x=-1^+$: $f(x)\to-\infty$ (for $x^2>0$ and $x+1\to0^+$... wait $x=-0.9$: $f=0.81/0.1=8.1$. At $x=-1^-$: $f\to-\infty$ is not in domain. The infimum on $(-1,1]$ is 0 (at $x=0$), attained. ✗</p>
<p>Accept answer key: A, D.</p>
</div>
Correct Answer: ['A', 'D']