Question

Решите пожалуйста,буду благодарен)срочно надо

Answer 1

$f'(x) = \frac{4}{3} ( x^{3})'-5(x ^{2})' + 6(x)' -9' = \frac{4}{3}*3 x^{2} -5*2x + 6=$$4 x^{2} -10x+6$
f '(x) = 0
4x² - 10x + 6= 0
2x² - 5x + 3 = 0
D = (- 5)² - 4 * 2 * 3 = 25 - 24 = 1
$x _{1} = \frac{5+1}{4}=1,5\\x _{2}= \frac{5-1}{4}=1$ * 3 = 25 - 24 = 1
Ответ: 1,5 ; 1

f(x) = 2x + 1                                           g(x) = 3x² + x
f ' (x) = 2(x)' + 1' = 2                              g ' (x) = 3(x²)' + (x)' = 6x + 1
                                  f ' (x) > g ' (x)
                                   2 > 6x + 1
                                      6x < 1
                                      x < 1/6
 Ответ : x ∈ (- ∞ ; 1/6)