Question

Y= x(x-4)^3 критически точки

Answer 1

найдем производную

y'=(x-4)^3+3x*(x-4)^2=(x-4)^2(x-4+3x)=(x-4)^2(4x-4)=4(x-1)(x-4)^2

x1=1   x2=4

x<1  y'<0

x>4  y'>0

x=1 точка минимума

y''=4*(x-4)^2+8*(x-1)(x-4)=(x-4)(4x-16+8x-8)=(x-4)(12x-24)=12*(x-4)(x-2)

x=4 x=2 точки перегиба

Answer 2

To find the critical points of the function $Y = x(x-4)^3$, we need to find the values of $x$ where the derivative of the function is equal to zero or undefined. Critical points occur when the derivative is zero or when the function is not differentiable at a certain point (undefined).

Let's start by finding the derivative of $Y$ with respect to $x$:

$Y = x(x-4)^3$

To find $Y'$, the derivative of $Y$ with respect to $x$, we can use the product rule and the chain rule:

$Y' = (x-4)^3 + x \cdot 3(x-4)^2 \cdot 1$

Now, set $Y'$ to zero and solve for $x$ to find the critical points:

$0 = (x-4)^3 + 3x(x-4)^2$

To solve this equation, let's factor out the common term $(x-4)^2$ from both terms:

$0 = (x-4)^2 \left[(x-4) + 3x\right]$

Now, let's simplify the expression inside the square brackets:

$0 = (x-4)^2(4x-4)$

Next, let's factor out the common term $4$ from the expression $4x-4$:

$0 = 4(x-1)(x-4)^2$

Now, we have two critical points:

1. When $x-1 = 0$, $x = 1$.
2. When $x-4 = 0$, $x = 4$.

So, the critical points of the function $Y = x(x-4)^3$ are $x = 1$ and $x = 4$.