Question

Решите уравнение f'(x)=0,если f(x)=x(x^2-5x+3)

Answer 1

$f(x)=x(x^2-5x+3)\\f(x)=x^3-5x^2+3x\\f'(x)=3x^2-10x+3\\f'(0)=0-0+3=3$

Answer 2

To find the critical points of the function $f(x) = x(x^2 - 5x + 3)$, we need to find the values of $x$ for which the derivative $f'(x)$ equals zero.

First, let's find the derivative of $f(x)$:

$f(x) = x(x^2 - 5x + 3)$

Using the product rule for differentiation, we get:

$f'(x) = x \frac{d}{dx}(x^2 - 5x + 3) + \frac{d}{dx}(x) (x^2 - 5x + 3)$

Now, let's calculate the derivatives:

$\frac{d}{dx}(x^2 - 5x + 3) = 2x - 5$ (derivative of $x^2$ is $2x$, derivative of $-5x$ is $-5$, derivative of a constant is $0$)

$\frac{d}{dx}(x) = 1$ (derivative of $x$ with respect to $x$ is $1$)

Substitute these derivatives back into $f'(x)$:

$f'(x) = x(2x - 5) + 1(x^2 - 5x + 3)$

$f'(x) = 2x^2 - 5x + x^2 - 5x + 3$

$f'(x) = 3x^2 - 10x + 3$

Now, we set $f'(x) = 0$ and solve for $x$:

$3x^2 - 10x + 3 = 0$

This is a quadratic equation, and we can solve it by factoring or using the quadratic formula. Factoring might not be straightforward, so let's use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 3$, $b = -10$, and $c = 3$:

$x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3}$

$x = \frac{10 \pm \sqrt{100 - 36}}{6}$

$x = \frac{10 \pm \sqrt{64}}{6}$

$x = \frac{10 \pm 8}{6}$

Now, we have two possible solutions for $x$:

1. $x = \frac{10 + 8}{6} = \frac{18}{6} = 3$
2. $x = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3}$

So, the critical points of the function $f(x) = x(x^2 - 5x + 3)$