Question

画图 $y = 2sqrt3{x^2}$ $y = x$

Explanation:

Step1: Analyze \( y = 2\sqrt[3]{x^2} \)

Rewrite \( y = 2\sqrt[3]{x^2} \) as \( y = 2x^{\frac{2}{3}} \). The domain is all real numbers (\( x\in\mathbb{R} \)). For \( x = 0 \), \( y = 0 \). As \( x\to\pm\infty \), \( y\to+\infty \) (since the exponent \( \frac{2}{3} \) is positive, and coefficient 2 is positive). The function is even? Wait, \( f(-x)=2(-x)^{\frac{2}{3}} = 2(x^2)^{\frac{1}{3}}=2x^{\frac{2}{3}}=f(x) \), so it's even? Wait, no: \( (-x)^{\frac{2}{3}} = [(-x)^2]^{\frac{1}{3}} = x^{\frac{2}{3}} \), so yes, even. So symmetric about y-axis. Let's find some points:

• \( x = -1 \): \( y = 2(-1)^{\frac{2}{3}} = 2(1) = 2 \)
• \( x = 1 \): \( y = 2(1)^{\frac{2}{3}} = 2 \)
• \( x = 8 \): \( y = 2(8)^{\frac{2}{3}} = 2(4) = 8 \) (since \( 8^{\frac{1}{3}} = 2 \), \( 2^2 = 4 \))
• \( x = -8 \): \( y = 2(-8)^{\frac{2}{3}} = 2(4) = 8 \)

Step2: Analyze \( y = x \)

This is a linear function with slope 1, y-intercept 0. Passes through (0,0), (1,1), (-1,-1), etc. It's symmetric about the origin (odd function: \( f(-x)=-x=-f(x) \)).

Step3: Plot Key Points

For \( y = 2x^{\frac{2}{3}} \):

• Points: (-8,8), (-1,2), (0,0), (1,2), (8,8)
• Shape: At \( x = 0 \), it has a "flatter" slope (derivative \( y' = 2\cdot\frac{2}{3}x^{-\frac{1}{3}}=\frac{4}{3x^{\frac{1}{3}}} \), undefined at \( x = 0 \), so a vertical tangent? Wait, as \( x\to0^+ \), \( y'\to+\infty \); as \( x\to0^- \), \( y'\to-\infty \), so a cusp at (0,0)? Wait, no, since it's even, left and right: for \( x>0 \), \( y'=\frac{4}{3x^{\frac{1}{3}}}>0 \), increasing; for \( x<0 \), \( y'=\frac{4}{3(-|x|)^{\frac{1}{3}}}<0 \), decreasing. So at x=0, the function changes from decreasing (x<0) to increasing (x>0), with a cusp? Wait, no, the derivative is undefined at x=0, but the function is continuous there.

For \( y = x \):

• Straight line through origin, slope 1.

Step4: Sketch the Graphs

• For \( y = 2\sqrt[3]{x^2} \): Plot the points (-8,8), (-1,2), (0,0), (1,2), (8,8). The curve is U-shaped (but symmetric about y-axis) with a cusp at (0,0) (since derivative is undefined there, and the function is smooth elsewhere).
• For \( y = x \): Draw a straight line through (0,0), (1,1), (-1,-1).

Answer:

To sketch \( y = 2\sqrt[3]{x^2} \) (a curve symmetric about the y - axis, passing through \((\pm1,2)\), \((\pm8,8)\), with a cusp at the origin) and \( y = x \) (a straight line through the origin with slope 1), plot the key points and draw the curves/lines as analyzed.