Question

Y=-2/3 x√x+3x+1 {1;9} max и min

Answer 1

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Answer 2

To find the maximum and minimum values of the function $y = -\frac{2}{3}x\sqrt{x} + 3x + 1$ in the interval $[1, 9]$, we need to find the critical points and evaluate the function at those points as well as at the endpoints of the interval.

1. First, let's find the critical points by taking the derivative of the function with respect to $x$ and setting it equal to zero:

$y = -\frac{2}{3}x\sqrt{x} + 3x + 1$

$y' = -\frac{2}{3}\sqrt{x} + \frac{3}{2} + 0$ (simplifying the derivative)

Setting $y'$ equal to zero:

$-\frac{2}{3}\sqrt{x} + \frac{3}{2} = 0$

$\frac{3}{2} = \frac{2}{3}\sqrt{x}$

Solving for $x$:

$\sqrt{x} = \frac{3}{2} \times \frac{3}{2}$

$\sqrt{x} = \frac{9}{4}$

$x = \left(\frac{9}{4}\right)^2$

$x = \frac{81}{16}$

So, we have found a critical point at $x = \frac{81}{16}$.

2. Next, we need to evaluate the function at the critical point and the endpoints of the interval.

For $x = 1$:

$y = -\frac{2}{3}(1)\sqrt{1} + 3(1) + 1 = -\frac{2}{3} + 3 + 1 = \frac{7}{3}$

For $x = \frac{81}{16}$:

$y = -\frac{2}{3}\left(\frac{81}{16}\right)\sqrt{\frac{81}{16}} + 3\left(\frac{81}{16}\right) + 1$ $= -\frac{2}{3}\left(\frac{81}{16}\right)\left(\frac{9}{4}\right) + \frac{243}{16} + 1$ $= -\frac{2}{3}\left(\frac{729}{64}\right) + \frac{243}{16} + 1$ $= -\frac{1458}{192} + \frac{243}{16} + 1$