Question

Решить неравенство f'x>0 если f(x)=(3x+4)(5-6x)

Answer 1

f(x)=(3x+4)(5-6x)=15x-18x²+20-24x=-18x²-9x+20

f'(x)=-36x-9

f'(x)>0

-36x-9>0

-36x>9

x<-1/4

Answer 2

f(x)=0

(3x+4)×(5-6x)=o

Система:

3x+4=0 3x=-4 x=-1,(3)

5-6x=0 -6x=-5 x= 5\6

Числовая ось....

И так тебе нужно указать знает

Идёт:-,+,-

Тебе нужно выбрать только то,что больше нулю

Значит:S=(-1,(3),5\6)

Answer 3

To solve the inequality $f'(x) > 0$ for the given function $f(x) = (3x + 4)(5 - 6x)$, we need to find the first derivative of $f(x)$ and determine the intervals where the derivative is positive.

Step 1: Find the first derivative $f'(x)$ of $f(x)$. To find the derivative, we'll use the product rule for differentiation: $f(x) = (3x + 4)(5 - 6x)$ $f'(x) = (3x + 4)'(5 - 6x) + (3x + 4)(5 - 6x)'$ $f'(x) = (3)(5 - 6x) + (3x + 4)(-6)$ $f'(x) = 15 - 18x - 18x - 24$ $f'(x) = -36x - 9$

Step 2: Find the critical points. To find the critical points, we set the first derivative equal to zero and solve for $x$: $f'(x) = -36x - 9 = 0$ $36x = -9$ $x = -\frac{9}{36} = -\frac{1}{4}$

Step 3: Determine the intervals where $f'(x) > 0$. Now, we need to identify the intervals where $f'(x)$ is positive to satisfy the inequality $f'(x) > 0$.

Since $f'(x) = -36x - 9$, we'll create a sign chart to determine the sign of $f'(x)$ in different intervals:

$$\begin{array}{|c|c|c|c|} \hline \text{Interval} & -\infty < x < -\frac{1}{4} & -\frac{1}{4} < x < \infty \\ \hline f'(x) & - & + \\ \hline \end{array}$$

From the sign chart, we can see that $f'(x) > 0$ in the interval $-\frac{1}{4} < x < \infty$.

Step 4: Write the final solution. Therefore, the solution to the inequality $f'(x) > 0$ for the function $f(x) = (3x + 4)(5 - 6x)$ is: $x \in \left(-\frac{1}{4}, \infty\right)$