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Ответ:

Дивись нижче

Объяснение:

1) f(x) = 8[tex]x^{2}[/tex]-13[tex]x^{3}[/tex]

f`(x) = 8*2x-13*3[tex]x^{2}[/tex]

f`(x) = 16x-39[tex]x^{2}[/tex]

16x-39[tex]x^{2}[/tex]=0

x1=0

x2=[tex]\frac{16}{39}[/tex]

f(0) = 0

f([tex]\frac{16}{39}[/tex]) = 8*  ([tex]\frac{16}{39}[/tex][tex])^{2}[/tex]-13[tex](\frac{16}{39}) ^{3}[/tex]= [tex]\frac{2048}{4563}[/tex]

fmax=   [tex]\frac{2048}{4563}[/tex]

fmin=0

2) f(x) =4[tex]x^{3[/tex]-13[tex]x^{2}[/tex]+8  

f`(x) = 4*3[tex]x^{2}[/tex]-13*2x+0

f`(x) = 12[tex]x^{2}[/tex]-26x

12[tex]x^{2}[/tex]-26x=0

2x(6x-13)=0

x(6x-13)=0

x=0

6x-13=0

x1=0

x2=[tex]\frac{13}{6}[/tex]

f(0) = 0

f([tex]\frac{13}{6}[/tex]) = 4*[tex](\frac{13}{6})^{3}[/tex]-13[tex](\frac{13}{6})^{2}[/tex]+8

f([tex]\frac{13}{6}[/tex]) = [tex]-\frac{1333}{108}[/tex]

fmax = 8

fmin= -8

[-8;8]

3. -