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 (\( x\in\mathbb{R} \)) since cube root is defined for all real \( x \), and squaring (or raising to \( \frac{2}{3} \)) is also defined for all real \( x \). For \( x = 0 \), \( y = 0 \). For positive \( x \), as \( x \) increases, \( x^{\frac{2}{3}} \) increases (since the exponent \( \frac{2}{3}>0 \)), so \( y \) increases. For negative \( x \), \( x^2 \) is positive, so \( y \) is positive (even function? Wait, \( f(-x)=2\sqrt[3]{(-x)^2}=2\sqrt[3]{x^2}=f(x) \), so it's even. So the graph is symmetric about the \( y \)-axis. Let's pick some points:

• \( x = -8 \): \( y = 2\sqrt[3]{64}=2\times4 = 8 \)
• \( x = -1 \): \( y = 2\sqrt[3]{1}=2\times1 = 2 \)
• \( x = 0 \): \( y = 0 \)
• \( x = 1 \): \( y = 2 \)
• \( x = 8 \): \( y = 8 \)

Step2: Analyze \( y = x \)

This is a linear function with slope \( 1 \) and \( y \)-intercept \( 0 \). The domain and range are all real numbers. Key points:

• \( x = -2 \), \( y = -2 \)
• \( x = 0 \), \( y = 0 \)
• \( x = 2 \), \( y = 2 \)

Step3: Sketch the graphs

• For \( y = 2\sqrt[3]{x^2} \): Plot the points \((-8,8)\), \((-1,2)\), \((0,0)\), \((1,2)\), \((8,8)\). Since it's even, reflect over \( y \)-axis. The curve is smooth, increasing for \( x\geq0 \), decreasing for \( x\leq0 \) (wait, no: for \( x\leq0 \), as \( x \) increases from \( -\infty \) to \( 0 \), \( x^2 \) decreases from \( +\infty \) to \( 0 \), so \( y = 2x^{\frac{2}{3}} \) decreases from \( +\infty \) to \( 0 \) when \( x \) goes from \( -\infty \) to \( 0 \), and increases from \( 0 \) to \( +\infty \) when \( x \) goes from \( 0 \) to \( +\infty \). Wait, earlier mistake: for \( x<0 \), as \( x \) increases (towards \( 0 \)), \( x^2 \) decreases, so \( y \) decreases? Wait no, \( x=-8 \) (small \( x \), more negative) gives \( y = 8 \), \( x=-1 \) (closer to \( 0 \)) gives \( y = 2 \), so as \( x \) increases from \( -\infty \) to \( 0 \), \( y \) decreases from \( +\infty \) to \( 0 \). Then from \( 0 \) to \( +\infty \), \( x \) increases, \( y \) increases from \( 0 \) to \( +\infty \). And symmetric about \( y \)-axis.

• For \( y = x \): Plot the points \((-2,-2)\), \((0,0)\), \((2,2)\) and draw a straight line with slope \( 1 \).

Now, sketch \( y = 2\sqrt[3]{x^2} \) as a curve symmetric about \( y \)-axis, passing through the points above, and \( y = x \) as a straight line through the origin with slope \( 1 \).

Answer:

To sketch \( y = 2\sqrt[3]{x^2} \) (symmetric about \( y \)-axis, curve through \((-8,8), (-1,2), (0,0), (1,2), (8,8)\)) and \( y = x \) (straight line through origin, slope \( 1 \)), follow the point - plotting and symmetry/linearity analysis above.