Răspuns:
Explicație pas cu pas:
a) Ec.tangentei:
y-y0 =f'(x0)(x -x0)
f' = e^x -2
f'(1) = e-2
y-1 =(e-2)x-1) ec.tangentei
b) f' =0
e^x -2 = 0
e^x = 2, x*lne = ln2, x = ln2
f' are o singura sol. , f are punct de extrem unic
c) g(x) = e^x -2x -1
g(1) = e -2 -1 = e -3 < 0 (e= 2,7...)
g(2) = e^2 -4 -1 = e^2 -5 > 0
g are o sol. in (1,2)
