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1.Sa se determine parametrul "a" astfel incat functia f sa fie continua in x0=1

f(x) = { 4xa-5 , x1

2. derivata :
a) f(x) = 3(xla a doua -4) la puterea 100
b) f= sin(7x-1)
c) f= 4e la puterea 5x-4 ori lnx


Răspuns :

raspunsul in word ,succes

[tex](4 e^{5x-4})' lnx+ 4e^{5x-4}*(lnx)'= 4 * e^{5x-4} *5x-4 + 4e^{5x-4} * \frac{1}{x} = 20x * e^{5x-4}-4+ 4e^{5x-4}* \frac{1}{x}= e^{5x-4} (20x+4+1/x) [/tex]

[tex]((3x^{2} -4)^{100})'= 100(3x^2-4)^{99} (6x) [/tex]

[tex]sin(7x-1)= cos(7x-1) *7[/tex]
Vezi imaginea ALESYO