Question

Решите неравенство f ' (x) < 0 f(x) = 12x^3 + 2x^4

Answer 1

решение задания смотри на фотографии е

решение задания смотри на фотографии

Answer 2

To solve the inequality $f'(x) < 0$ for the function $f(x) = 12x^3 + 2x^4$, we need to find the critical points and determine the intervals where the derivative is negative.

Step 1: Find the derivative $f'(x)$ of the function $f(x)$: $f'(x) = \frac{d}{dx} (12x^3 + 2x^4) = 36x^2 + 8x^3.$

Step 2: Find the critical points by setting the derivative equal to zero and solving for $x$: $36x^2 + 8x^3 = 0.$

Step 3: Factor out $x^2$ from the equation: $x^2(36 + 8x) = 0.$

Now, we have two critical points: $x = 0$ and $36 + 8x = 0$. Solving for $x$ in the second equation: $36 + 8x = 0 \implies 8x = -36 \implies x = -\frac{36}{8} \implies x = -\frac{9}{2}.$

Step 4: Create a sign chart to determine the intervals where $f'(x)$ is negative:

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       -∞     -9/2     0       +∞
       --------------------------
f'(x)  |  +     0     -     0     +

Step 5: Analyze the results:

The sign chart shows that $f'(x)$ is negative in the interval $(-9/2, 0)$.

Step 6: Write the solution to the inequality:

The solution to $f'(x) < 0$ for the function $f(x) = 12x^3 + 2x^4$ is $x \in \left(-\frac{9}{2}, 0\right)$.