Question

Найдите значения параметра m, при которых функция f(x)=x^3+3mx^2+5mx возрастает на R.

Answer 1

Функция возрастает на R, если ее производная на всей R >0.
найдем производную
f'(x)=3x²+6mx+5m
квадратный трехчлен будет больше нуля для любых х, если его дискриминант будет <0.
найдем дискриминант
Д=(6m)²-4*3*5m=36m²-60m
36m²-60m<0
12m(3m-5)<0 |:12
m(3m-5)<0
нули: m=0, m=5/3
mє(0;5/3)

Answer 2

For the function f(x) = x^3 + 3mx^2 + 5mx to be increasing (or strictly increasing) on the entire real number line R, the first derivative f'(x) must be positive for all x in R.

Let's find the first derivative of f(x) with respect to x:

f(x) = x^3 + 3mx^2 + 5mx

f'(x) = d/dx (x^3 + 3mx^2 + 5mx) f'(x) = 3x^2 + 6mx + 5m

Now, for the function to be increasing on R, f'(x) > 0 for all x in R.

3x^2 + 6mx + 5m > 0

To ensure this inequality holds for all x in R, we can consider the discriminant of the quadratic expression:

Discriminant (D) = b^2 - 4ac where a = 3, b = 6m, and c = 5m

D = (6m)^2 - 4 * 3 * 5m D = 36m^2 - 60m

For the function to be increasing on R, the discriminant D must be less than zero (D < 0). This ensures that the quadratic expression 3x^2 + 6mx + 5m is always positive, and the function is strictly increasing on R.

So, we have:

36m^2 - 60m < 0

To solve this inequality, we can factor out m:

m(36m - 60) < 0

Now, we need to find the critical points where the inequality might change its sign. These points are where m = 0 and where 36m - 60 = 0.

1. m = 0
2. 36m - 60 = 0 36m = 60 m = 60/36 m = 5/3 ≈ 1.67

Now, let's check the intervals between these critical points and the sign of the inequality in each interval:

1. For m < 0: Choose m = -1, for example. m(36m - 60) = (-1)(36(-1) - 60) = (-1)(-96) = 96 > 0

2. For 0 < m < 5/3: Choose m = 1, for example. m(36m - 60) = (1)(36(1) - 60) = (1)(-24) = -24 < 0

3. For m > 5/3: Choose m = 2, for example. m(36m - 60) = (2)(36(2) - 60) = (2)(12) = 24 > 0

The sign changes from positive to negative at m = 5/3 ≈ 1.67. Thus, the function f(x) = x^3 + 3mx^2 + 5mx is strictly increasing for m < 5/3 (or m ∈ (-∞, 5/3)).