jacks said:
mathbalarka said:
| Solving for the number of real roots of \(\displaystyle f(x) = x_1=root_1\) means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_1$, as shown by the green line and we see that in this case we have only one real root for \(\displaystyle f(x) = x_1=root_1\). | Solving for the number of real roots of \(\displaystyle f(x) = x_2=root_2\) means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_2$, as shown by the purple line and we see that in this case we have only one real root for \(\displaystyle f(x) = x_2=root_2\). | Solving for the number of real roots of \(\displaystyle f(x) = x_3=root_3\) means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_3$, as shown by the orange line and we see that in this case we have only one real root for \(\displaystyle f(x) = x_3=root_3\). |
anemone said:
mathbalarka said:
| $f(x)=x^3-3x+1$ | $g(x)=f(f(x))=(f(x))^3-3(f(x))+1$ |
| $f'(x)=3x^2-3=3(x^2-1)=3(x+1)(x-1)$ $\rightarrow f'(x)=0$ iff $3(x+1)(x-1)=0$ or $x=\pm1$. | $g'(x)=3f(x)^2f'(x)-3f'(x)=3f'(x)((f(x))^2-1)$ $\rightarrow g'(x)=0$ iff $x=\pm1$, $x=\pm \sqrt{3}$, or $x=-2$ or $x=0$. |
| $f''(x)=6x$ | $g''(x)=6f(x)(f'(x))^2+3f''(x)((f(x))^2-1)$ |
| $x$ | -2 | $-\sqrt{3}$ | -1 | 0 | 1 | $\sqrt{3}$ |
| $f(x)$ | -1 | 1 | 3 | 1 | -1 | 1 |
| $f'(x)$ | 9 | 6 | 0 | -3 | 0 | 6 |
| $f''(x)$ | -12 | $-6\sqrt{3}$ | -6 | 0 | 6 | $6\sqrt{3}$ |
| $g(x)$ | 3 | -1 | 19 | -1 | 3 | -1 |
| $g'(x)$ | 0 | 0 | 0 | 0 | 0 | 0 |
| $g''(x)$ | -486 | 216 | -144 | 54 | 0 | 216 |
mathbalarka said:
Distinct real roots refer to the number of unique solutions that a polynomial equation has when solved for the variable. These solutions must be real numbers, meaning they are not complex or imaginary.
To find the number of distinct real roots, you can use the fundamental theorem of algebra, which states that a polynomial equation of degree n will have n distinct complex roots. To find the number of distinct real roots, you can use synthetic division or the rational root theorem to determine the number of real roots before solving for their values.
Yes, a polynomial equation can have multiple distinct real roots. The number of distinct real roots is equal to the degree of the polynomial, but it is possible for some of these roots to be repeated. For example, a quadratic equation can have two distinct real roots, but they can also both be the same value.
Distinct real roots are unique solutions to a polynomial equation, while repeated real roots are solutions that occur more than once. In other words, distinct real roots are different values, while repeated real roots are the same value. For example, the equation x^2 - 4x + 4 = 0 has two repeated real roots of x = 2, while the equation x^2 - 5x + 6 = 0 has two distinct real roots of x = 2 and x = 3.
Yes, it is possible for a polynomial equation to have no distinct real roots. This occurs when all of the solutions are complex or imaginary numbers. For example, the equation x^2 + 1 = 0 has no real roots, as the solutions are complex numbers (±i).
