Analyse the extreme curve sketching of the function f(x)=x3−3x.

The x-coordinate of the vertex for a parabola $f(x) = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$.
For $f(x) = x^2 - 3x$, we have $a=1$ and $b=-3$.

$$x = -\frac{-3}{2(1)} = \frac{3}{2}$$

Now, substitute $x=\frac{3}{2}$ into the original function to find the y-coordinate:

$$f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right)$$ $$f\left(\frac{3}{2}\right) = \frac{9}{4} - \frac{9}{2}$$ $$f\left(\frac{3}{2}\right) = \frac{9}{4} - \frac{18}{4}$$ $$f\left(\frac{3}{2}\right) = -\frac{9}{4}$$

The vertex (extreme point) is $\left(\frac{3}{2}, -\frac{9}{4}\right)$. Since the coefficient of $x^2$ is positive ($a=1$), this vertex is a local minimum.

To find the y-intercept, set $x=0$:

$$f(0) = (0)^2 - 3(0) = 0$$

The y-intercept is $(0, 0)$.

To find the x-intercepts, set $f(x)=0$:

$$x^2 - 3x = 0$$

Factor out $x$:

$$x(x - 3) = 0$$

This gives two x-intercepts:

$$x = 0 \quad \text{or} \quad x - 3 = 0 \implies x = 3$$

The x-intercepts are $(0, 0)$ and $(3, 0)$.

Find the first derivative of $f(x)$:

$$f'(x) = \frac{d}{dx}(x^2 - 3x) = 2x - 3$$

The function is increasing when $f'(x) > 0$:

$$2x - 3 > 0 \implies 2x > 3 \implies x > \frac{3}{2}$$

So, $f(x)$ is increasing on the interval $\left(\frac{3}{2}, \infty\right)$.
The function is decreasing when $f'(x) < 0$:

$$2x - 3 < 0 \implies 2x < 3 \implies x < \frac{3}{2}$$

So, $f(x)$ is decreasing on the interval $\left(-\infty, \frac{3}{2}\right)$.

Find the second derivative of $f(x)$:

$$f''(x) = \frac{d}{dx}(2x - 3) = 2$$

Since $f''(x) = 2 > 0$ for all $x$, the function is concave up everywhere. There are no inflection points.

Summary of analysis for curve sketching:

• The function is a parabola opening upwards.
• The vertex (local minimum) is at $\boxed{\left(\frac{3}{2}, -\frac{9}{4}\right)}$.
• The y-intercept is at $\boxed{(0, 0)}$.
• The x-intercepts are at $\boxed{(0, 0) \text{ and } (3, 0)}$.
• The function is decreasing on $\boxed{\left(-\infty, \frac{3}{2}\right)}$ and increasing on $\boxed{\left(\frac{3}{2}, \infty\right)}$.
• The function is concave up on $\boxed{(-\infty, \infty)}$.