Question

画图 $y = 2sqrt3{x^2}$ $y = x$

Explanation:

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

The function \( y = x^{\frac{2}{3}} \) is a power function. For \( y = 2x^{\frac{2}{3}} \), the domain is all real numbers (\( x\in\mathbb{R} \)) since we can take the cube - root of any real number, and then square it. The range is \( y\geq0 \) because \( x^{2}\geq0 \), so \( \sqrt[3]{x^{2}}\geq0 \), and multiplying by 2 (a positive number) still gives \( y\geq0 \). The function is even? Wait, no, \( f(-x)=2\sqrt[3]{(-x)^{2}} = 2\sqrt[3]{x^{2}}=f(x) \), so it is even, symmetric about the y - axis. Let's find some key points:

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

Step2: Analyze \( y = x \)

The function \( y = x \) is a linear function with a slope of 1 and a y - intercept of 0. The domain and range are all real numbers (\( x\in\mathbb{R},y\in\mathbb{R} \)). Key points:

• When \( x = 0 \), \( y = 0 \).
• When \( x = 1 \), \( y = 1 \).
• When \( x=-1 \), \( y=-1 \).

Step3: Sketch the graphs

• For \( y = 2\sqrt[3]{x^{2}} \):
• Plot the points we found: (0,0), (1,2), (- 1,2), (8,8), (-8,8). Since the function is even, we can draw the right - hand side (for \( x\geq0 \)) and then reflect it over the y - axis to get the left - hand side. The graph for \( x\geq0 \) is increasing (we can check the derivative: \( y^\prime=2\times\frac{2}{3}x^{-\frac{1}{3}}=\frac{4}{3x^{\frac{1}{3}}} \), for \( x > 0 \), \( y^\prime>0 \), so it is increasing on \( (0,+\infty) \); for \( x < 0 \), \( y^\prime<0 \), so it is decreasing on \( (-\infty,0) \)) and has a "cup - like" shape near the origin (but flatter than a parabola) and becomes steeper as \( |x| \) increases.
• For \( y=x \):
• Plot the points (0,0), (1,1), (-1,-1). Then draw a straight line with a slope of 1 passing through these points. The line \( y = x \) passes through the origin and makes a \( 45^{\circ} \) angle with the x - axis.

Answer:

To sketch \( y = 2\sqrt[3]{x^{2}} \) (a non - negative, even, power - function - based curve) and \( y = x \) (a straight line with slope 1), follow the steps of analyzing domain, range, symmetry, key points, and then plotting and connecting the points. The graph of \( y = 2\sqrt[3]{x^{2}} \) is a curve symmetric about the y - axis, passing through (0,0), (1,2), (-1,2), etc., and the graph of \( y = x \) is a straight line passing through the origin with a slope of 1.