Question

Функцию задано формулой f(x) = 3x² +2x. Найдите:1) f(1); f(0); f(⅓); f(-2)2) найдите значение аргумента, при котором значение функции равно:0; -1; -56.3) есть ли правильным равенство:f (-1) = 5;f(2)= -8?

Answer 1

$f(x)=3x^{2} +2x=x(3x+2)$

1:

• $f(1) = 1*(3*1+2) = 6$
• 0
• $f(\frac{1}{3}) = \frac{1}{3} (1+2) = 1$
• 8

2:

1. $f(x) = 0 = x(3x+2)$ ,   x = 0
2. $3x^2+2x + 1 = 0$, нет решений
3. $3x^2+2x+56=0$,  нет решений

3:

• $f(-1) = 3 - 2 \neq 5$
• $f(2) = 3 * 4 + 4 \neq -8$

Answer 2

To find the values of the function $f(x) = 3x^2 + 2x$, we substitute the given values of $x$ into the function.

1. Calculate the values of the function for different $x$ values: a) $f(1) = 3 \cdot 1^2 + 2 \cdot 1 = 3 + 2 = 5$ b) $f(0) = 3 \cdot 0^2 + 2 \cdot 0 = 0 + 0 = 0$ c) $f(\frac{1}{3}) = 3 \cdot \left(\frac{1}{3}\right)^2 + 2 \cdot \frac{1}{3} = 3 \cdot \frac{1}{9} + \frac{2}{3} = \frac{1}{3} + \frac{2}{3} = 1$ d) $f(-2) = 3 \cdot (-2)^2 + 2 \cdot (-2) = 3 \cdot 4 - 4 = 12 - 4 = 8$

2. Find the values of $x$ for which $f(x)$ equals 0, -1, and -56. a) $f(x) = 0$ $3x^2 + 2x = 0$ Factor out common terms: $x(3x + 2) = 0$ So, $x = 0$ or $x = -\frac{2}{3}$

   b) $f(x) = -1$ $3x^2 + 2x = -1$ $3x^2 + 2x + 1 = 0$ This is a quadratic equation with no real solutions, so there are no real values of $x$ for which $f(x) = -1$.

   c) $f(x) = -56$ $3x^2 + 2x = -56$ $3x^2 + 2x + 56 = 0$ This is a quadratic equation with no real solutions, so there are no real values of $x$ for which $f(x) = -56$.

3. Check whether the given equalities hold true: a) $f(-1) = 5$ $f(-1) = 3 \cdot (-1)^2 + 2 \cdot (-1) = 3 \cdot 1 - 2 = 3 - 2 = 1$ The equality $f(-1) = 5$ is incorrect.

   b) $f(2) = -8$ $f(2) = 3 \cdot 2^2 + 2 \cdot 2 = 3 \cdot 4 + 4 = 12 + 4 = 16$ The equality $f(2) = -8$ is incorrect.

To summarize:

1. $f(1) = 5$, $f(0) = 0$, $f(\frac{1}{3}) = 1$, $f(-2) = 8$
2. For $f(x) = 0$, $x = 0$ or $x = -\frac{2}{3}$ For $f(x) = -1$ and $f(x) = -56$, there are no real solutions.
3. The given equalities $f(-1) = 5$