Question

1.f'(x)=0 f(x)=10√x-x+3 2.f'(x)>0 f(x)=x^3-3x+7

Answer 1

Решение смотри ниже.

Answer 2

To find the critical points of the functions and analyze their increasing or decreasing behavior, we'll need to calculate the first derivative for each function and then examine its sign.

1. $f(x) = 10\sqrt{x} - x + 3$:

Let's first find the first derivative, $f'(x)$, and then solve for critical points by setting $f'(x) = 0$.

$f(x) = 10\sqrt{x} - x + 3$

Taking the derivative:

$f'(x) = \frac{d}{dx} \left(10\sqrt{x} - x + 3\right)$

Using the power rule and chain rule:

$f'(x) = 10 \cdot \frac{1}{2} x^{-\frac{1}{2}} - 1$

Simplify:

$f'(x) = 5x^{-\frac{1}{2}} - 1$

Setting $f'(x) = 0$ to find critical points:

$5x^{-\frac{1}{2}} - 1 = 0$

Solving for $x$:

$5x^{-\frac{1}{2}} = 1$

$x^{-\frac{1}{2}} = \frac{1}{5}$

Taking the reciprocal of both sides:

$x^{\frac{1}{2}} = 5$

$x = 25$

Now, we can analyze the sign of $f'(x)$ around the critical point $x = 25$ to determine the increasing or decreasing behavior.

When $x < 25$, $f'(x) > 0$ (positive) When $x > 25$, $f'(x) < 0$ (negative)

This indicates that $f(x)$ is increasing to the left of $x = 25$ and decreasing to the right of $x = 25$, implying a local maximum at $x = 25$.

2. $f(x) = x^3 - 3x + 7$:

Let's find the first derivative, $f'(x)$, and then examine its sign.

$f(x) = x^3 - 3x + 7$

Taking the derivative:

$f'(x) = \frac{d}{dx} \left(x^3 - 3x + 7\right)$

Using the power rule:

$f'(x) = 3x^2 - 3$

Setting $f'(x) > 0$ to determine where the function is increasing:

$3x^2 - 3 > 0$

Dividing both sides by 3:

$x^2 - 1 > 0$

Factoring:

$(x + 1)(x - 1) > 0$

We have two intervals to consider: $(- \infty, -1)$ and $(-1, 1)$. In both intervals, the product of the factors is positive, so $f'(x) > 0$ in these intervals. This means that $f(x)$ is increasing in the intervals $(- \infty, -1)$ and $(-1, 1)$.

Beyond $x = 1$