Question

画图 y=2\sqrt3{x^2} y=x

Explanation:

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

First, rewrite the function in exponent form: \( y = 2x^{\frac{2}{3}} \). The domain of this function is all real numbers (\( x\in\mathbb{R} \)) because the cube root is defined for all real numbers, and squaring a real number is also defined for all real numbers. The function is even? Wait, no, \( f(-x)=2(-x)^{\frac{2}{3}} = 2((-x)^2)^{\frac{1}{3}}=2(x^2)^{\frac{1}{3}} = f(x) \), so it is even. So the graph is symmetric about the y - axis.

Step2: Find key points

• When \( x = 0 \), \( y=2\times0^{\frac{2}{3}} = 0 \), so the point (0,0) is on the graph.
• When \( x = 1 \), \( y = 2\times1^{\frac{2}{3}}=2\times1 = 2 \), so the point (1,2) is on the graph.
• When \( x=- 1 \), since the function is even, \( y = 2\times(-1)^{\frac{2}{3}}=2\times((-1)^2)^{\frac{1}{3}}=2\times1^{\frac{1}{3}} = 2 \), so the point (- 1,2) is on the graph.
• When \( x = 8 \), \( y=2\times8^{\frac{2}{3}}=2\times(2^3)^{\frac{2}{3}}=2\times2^{2}=2\times4 = 8 \), so the point (8,8) is on the graph.
• When \( x=-8 \), \( y = 2\times(-8)^{\frac{2}{3}}=2\times((-8)^2)^{\frac{1}{3}}=2\times64^{\frac{1}{3}}=2\times4 = 8 \), so the point (-8,8) is on the graph.

Step3: Analyze the shape

The function \( y = x^{\frac{2}{3}} \) has a shape that is similar to a parabola but with a flatter curve near the origin and steeper as \( |x| \) increases. Multiplying by 2 vertically stretches the graph of \( y = x^{\frac{2}{3}} \) by a factor of 2.

Step4: Compare with \( y = x \)

The line \( y=x \) is a straight line with a slope of 1 passing through the origin. For \( x>0 \), we can compare the values:

• At \( x = 1 \), \( y = 2\sqrt[3]{1^2}=2 \) and \( y=x = 1 \), so \( 2\sqrt[3]{x^2}>x \) when \( x = 1 \).
• At \( x = 0 \), both \( y = 2\sqrt[3]{0^2}=0 \) and \( y=x = 0 \), so they intersect at (0,0).
• Let's find the intersection points of \( y = 2\sqrt[3]{x^2} \) and \( y=x \) (for \( x\geq0 \), since for \( x < 0 \), \( y = 2\sqrt[3]{x^2}>0 \) and \( y=x<0 \), so no intersection for \( x < 0 \)):

Set \( 2x^{\frac{2}{3}}=x \). Let \( t=x^{\frac{1}{3}} \), then \( x = t^3 \). The equation becomes \( 2t^{2}=t^3 \), \( t^3 - 2t^{2}=0 \), \( t^{2}(t - 2)=0 \). So \( t = 0 \) or \( t = 2 \). When \( t = 0 \), \( x = 0 \); when \( t = 2 \), \( x=t^3=8 \). So the graphs intersect at (0,0) and (8,8).

To draw \( y = 2\sqrt[3]{x^2} \):

1. Plot the points we found: (0,0), (1,2), (-1,2), (8,8), (-8,8).
2. Since the function is even, reflect the part for \( x\geq0 \) over the y - axis to get the part for \( x\leq0 \).
3. The curve is smooth, passing through these points, symmetric about the y - axis, with a flatter curve near the origin and getting steeper as \( |x| \) increases.

The line \( y = x \) is a straight line with slope 1, passing through (0,0), (1,1), (2,2), etc.

Answer:

To draw \( y = 2\sqrt[3]{x^2} \) (rewritten as \( y = 2x^{\frac{2}{3}} \)):

1. Symmetry: The function is even (\( f(-x)=f(x) \)), so the graph is symmetric about the \( y \) - axis.
2. Key Points:

• At \( x = 0 \), \( y = 0 \) (intersection with \( y=x \)).
• At \( x=\pm1 \), \( y = 2 \).
• At \( x=\pm8 \), \( y = 8 \) (intersection with \( y=x \) for \( x = 8 \), and by symmetry for \( x=-8 \)).

1. Shape: It is a smooth curve, flatter near the origin, steeper as \( |x| \) increases, and vertically stretched (by a factor of 2) from the basic \( y=x^{\frac{2}{3}} \) curve.

To draw \( y = x \):

1. It is a straight line with a slope of 1, passing through the origin \((0,0)\) and points like \((1,1)\), \((2,2)\), \((8,8)\) (where it intersects \( y = 2\sqrt[3]{x^2} \)).

The graph of \( y = 2\sqrt[3]{x^2} \) is a symmetric (about \( y \) - axis) curve intersecting \( y = x \) at \((0,0)\) and \((8,8)\), with \( y = 2\sqrt[3]{x^2}\geq|x|/2\) (qualitatively) and the line \( y = x \) is a straight line with slope 1.