Question

1) f(x) = 9x^3-(2)/(x^3)+4(√x) -0,12 2)f(x)= (3-x^2)/(2x-1); f'(-1)

Answer 1

Вопрос то какой? Найти нули функции?

Answer 2

Let's solve each part step-by-step:

1. Given function: f(x) = 9x^3 - (2)/(x^3) + 4√x - 0.12

To find the derivative, f'(x), we'll differentiate each term with respect to x using the rules of differentiation:

a) Differentiating 9x^3 with respect to x: d/dx (9x^3) = 27x^2

b) Differentiating -2/x^3 with respect to x: d/dx (-2/x^3) = -(d/dx (2/x^3)) = -(2 * d/dx (x^(-3))) = -(2 * (-3x^(-4))) = 6/x^4

c) Differentiating 4√x with respect to x: d/dx (4√x) = 4 * (d/dx (x^(1/2))) = 4 * (1/2) * x^(-1/2) = 2/x^(1/2) = 2/√x

Now, let's put it all together:

f'(x) = 27x^2 + 6/x^4 + 2/√x

2. Given function: f(x) = (3 - x^2) / (2x - 1)

To find f'(-1), we need to evaluate the derivative of f(x) at x = -1. We'll follow similar steps as before:

a) Differentiating (3 - x^2) with respect to x: d/dx (3 - x^2) = -2x

b) Differentiating (2x - 1) with respect to x: d/dx (2x - 1) = 2

Now, let's evaluate f'(-1):

f'(-1) = -2(-1) / 2 = 2/2 = 1

So, the value of f'(-1) is 1.