Question

画图 \\( \

$$\begin{cases} y = 2\\sqrt3{x^2} \\\\ y = x \\end{cases}$$

\\)

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 \( \frac{2}{3} \) makes it even - like in behavior for sign, and the coefficient 2 scales it). The function is even? Wait, \( f(-x)=2(-x)^{\frac{2}{3}} = 2(x^{\frac{2}{3}})=f(x) \), so it's symmetric about the y - axis. Let's find some points: when \( x = 1 \), \( y = 2(1)^{\frac{2}{3}}=2 \); when \( x=- 1 \), \( y = 2(-1)^{\frac{2}{3}}=2((-1)^2)^{\frac{1}{3}} = 2(1)^{\frac{1}{3}} = 2 \); when \( x = 8 \), \( y=2(8)^{\frac{2}{3}}=2(2^3)^{\frac{2}{3}}=2\times4 = 8 \); when \( x=-8 \), \( y = 2(-8)^{\frac{2}{3}}=2((-8)^2)^{\frac{1}{3}}=2(64)^{\frac{1}{3}}=2\times4 = 8 \).

Step2: Analyze \( y = x \)

This is a linear function with slope \( m = 1 \) and y - intercept \( b = 0 \). The domain and range are all real numbers. Some points: when \( x = 0 \), \( y = 0 \); when \( x = 1 \), \( y = 1 \); when \( x=-1 \), \( y=-1 \).

Step3: Find intersection points

Set \( 2x^{\frac{2}{3}}=x \). Let \( t = x^{\frac{1}{3}} \), then \( x=t^3 \). The equation becomes \( 2t^2=t^3 \), or \( t^3 - 2t^2=0 \), \( t^2(t - 2)=0 \). So \( t = 0 \) or \( t = 2 \). When \( t = 0 \), \( x = 0 \), \( y = 0 \). When \( t = 2 \), \( x=t^3 = 8 \), \( y = 8 \). Also, we can check \( t\) negative: if \( t=-k,k>0 \), then the left - hand side of \( 2t^2=t^3 \) is \( 2k^2 \) (positive) and the right - hand side is \( -k^3 \) (negative), so no negative \( t \) solutions. So the curves intersect at \( (0,0) \) and \( (8,8) \).

Step4: Sketch the graphs

• For \( y = 2x^{\frac{2}{3}} \): Plot the points we found (\( (-8,8),(-1,2),(0,0),(1,2),(8,8) \)) and draw a smooth curve symmetric about the y - axis, increasing for \( x>0 \) (since the derivative \( y^\prime=2\times\frac{2}{3}x^{-\frac{1}{3}}=\frac{4}{3x^{\frac{1}{3}}} \), for \( x > 0 \), \( y^\prime>0 \); for \( x<0 \), \( y^\prime<0 \), so decreasing on \( (-\infty,0) \) and increasing on \( (0,+\infty) \))
• For \( y=x \): Draw a straight line with slope 1 passing through the origin, passing through points like \( (1,1) \), \( (8,8) \) etc.

Answer:

To sketch the graphs of \( y = 2\sqrt[3]{x^2} \) (or \( y = 2x^{\frac{2}{3}} \)) and \( y = x \):

1. Graph of \( y = 2x^{\frac{2}{3}} \):

• Domain: \( \mathbb{R} \), symmetric about the y - axis.
• Key points: \( (-8,8),(-1,2),(0,0),(1,2),(8,8) \).
• Behavior: Decreasing on \( (-\infty,0) \), increasing on \( (0,+\infty) \), \( y\to+\infty \) as \( x\to\pm\infty \).

1. Graph of \( y = x \):

• A straight line with slope 1, passing through the origin, with points like \( (0,0),(1,1),(8,8) \).

1. Intersection Points:

• The two graphs intersect at \( (0,0) \) and \( (8,8) \).

When sketching, draw the curve for \( y = 2x^{\frac{2}{3}} \) (smooth, symmetric about y - axis, passing through the key points) and the straight line \( y = x \) (straight, slope 1) on the same coordinate system.