Function Comparison: $f(x)$ vs $g(x)$

We are given two functions: $f(x) = x\sin(x)$ and $g(x)=|x|\sin(x)$. We need to determine the correct statement among the given options.

Analysis of Options

Option 1: $g(x) = |f(x)|$?

Let's compare $g(x)$ and $|f(x)|$. $g(x) = |x|\sin(x)$ $|f(x)| = |x\sin(x)|$ These are not always equal. For example, consider $x = \frac{3\pi}{2}$. $f(\frac{3\pi}{2}) = \frac{3\pi}{2}\sin(\frac{3\pi}{2}) = \frac{3\pi}{2}(-1) = -\frac{3\pi}{2}$. $|f(\frac{3\pi}{2})| = |-\frac{3\pi}{2}| = \frac{3\pi}{2}$. $g(\frac{3\pi}{2}) = |\frac{3\pi}{2}|\sin(\frac{3\pi}{2}) = \frac{3\pi}{2}(-1) = -\frac{3\pi}{2}$. Since $g(\frac{3\pi}{2}) \neq |f(\frac{3\pi}{2})|$, this statement is false.

Option 2: $g(x)$ is an even function?

A function $h(x)$ is even if $h(-x) = h(x)$ for all $x$. Let's check $g(x)$: $g(-x) = |-x|\sin(-x)$ Since $|-x| = |x|$ and $\sin(-x) = -\sin(x)$, we have: $g(-x) = |x|(-\sin(x)) = -|x|\sin(x) = -g(x)$. Because $g(-x) = -g(x)$, the function $g(x)$ is an odd function, not an even function. Thus, this statement is false.

Option 3: Identical local maxima x-coordinates?

Local maxima occur where the first derivative is zero and the second derivative is negative. For $f(x) = x\sin(x)$, $f'(x) = \sin(x) + x\cos(x)$. Maxima occur when $f'(x)=0$ and $f''(x)<0$. $f'(x)=0 \implies \tan(x) = -x$. For $g(x) = |x|\sin(x)$: If $x > 0$, $g(x) = x\sin(x)$. $g'(x) = \sin(x) + x\cos(x)$. Maxima occur when $g'(x)=0$ and $g''(x)<0$. $g'(x)=0 \implies \tan(x) = -x$. If $x < 0$, $g(x) = -x\sin(x)$. $g'(x) = -(\sin(x) + x\cos(x))$. Maxima occur when $g'(x)=0$ and $g''(x)<0$. $g'(x)=0 \implies \tan(x) = -x$. However, the second derivative test differs. For negative $x$ values satisfying $\tan(x)=-x$, let $x_0 < 0$. We found $f''(x_0) = \cos(x_0)(2+x_0^2)$. Since $x_0$ is in Q4, $\cos(x_0)>0$, so $f''(x_0)>0$, indicating a local minimum for $f(x)$. But for $g(x)$, we found $g''(x_0) = -\cos(x_0)(2+x_0^2) < 0$, indicating a local maximum for $g(x)$. Since the nature of the critical points differs, the x-coordinates for local maxima are not identical. This statement is false.

Option 4: $g(x)$ is differentiable at $x=0$?

We use the limit definition of the derivative: $g'(0) = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h}$ $g(0) = |0|\sin(0) = 0$. $g'(0) = \lim_{h \to 0} \frac{|h|\sin(h) - 0}{h} = \lim_{h \to 0} \frac{|h|\sin(h)}{h}$ Evaluate the limits from the left and right:

• Right-hand limit ($h \to 0^+$): $\lim_{h \to 0^+} \frac{|h|\sin(h)}{h} = \lim_{h \to 0^+} \frac{h\sin(h)}{h} = \lim_{h \to 0^+} \sin(h) = \sin(0) = 0$.
• Left-hand limit ($h \to 0^-$): $\lim_{h \to 0^-} \frac{|h|\sin(h)}{h} = \lim_{h \to 0^-} \frac{-h\sin(h)}{h} = \lim_{h \to 0^-} -\sin(h) = -\sin(0) = 0$.

Since the left-hand limit equals the right-hand limit, the derivative exists at $x=0$, and $g'(0) = 0$. This statement is true.

Conclusion

The only true statement is that $g(x)$ is differentiable at $x = 0$.