Question

Найти наибольшее и наименьшее значения функции f(x)=x³+3x²-72x+90 на отрезке [-4;5]

Answer 1

D/dx или иначе f'(x)=3x²+6x-72

d/dx(x^3)=3x²
d/dx(3x²)=6x
d/dx(-72x)=-72
d/dx(90)=0; (производная любого числа равно нулю)

затем приравниваем к нулю:
3x²+6x-72=0
x1,2=(b/2+-√(b/2²-ac))/a
x1=-6
x2=4

4-наибольшее

Answer 2

To find the maximum and minimum values of the function $f(x)=x^3+3x^2-72x+90$ on the interval $[-4, 5]$, we need to check the critical points and endpoints of the interval.

Step 1: Find the critical points Critical points are the points where the derivative of the function is equal to zero or undefined. To find the critical points, we take the derivative of $f(x)$ and set it equal to zero:

$f'(x) = 3x^2 + 6x - 72$

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

$3x^2 + 6x - 72 = 0$

Step 2: Solve for $x$

To solve this quadratic equation, we can factor it:

$3x^2 + 6x - 72 = 3(x^2 + 2x - 24) = 3(x + 6)(x - 4) = 0$

Setting each factor to zero and solving for $x$, we get:

$x + 6 = 0 \Rightarrow x = -6$

$x - 4 = 0 \Rightarrow x = 4$

So, the critical points are $x = -6$ and $x = 4$.

Step 3: Evaluate the function at the critical points and endpoints

Now, we need to evaluate the function at the critical points and the endpoints of the interval $[-4, 5]$:

$f(-4) = (-4)^3 + 3(-4)^2 - 72(-4) + 90 = -64 + 48 + 288 + 90 = 362$

$f(-6) = (-6)^3 + 3(-6)^2 - 72(-6) + 90 = -216 + 108 + 432 + 90 = 414$

$f(4) = 4^3 + 3(4)^2 - 72(4) + 90 = 64 + 48 - 288 + 90 = -86$

$f(5) = 5^3 + 3(5)^2 - 72(5) + 90 = 125 + 75 - 360 + 90 = -70$

Step 4: Determine the maximum and minimum values

The maximum value will be the highest of the values obtained, and the minimum value will be the lowest of the values obtained:

The maximum value is 414, which occurs at $x = -6$.

The minimum value is -86, which occurs at $x = 4$.

So, the maximum value of the function on the interval $[-4, 5]$ is 414, and the minimum value is -86.