Given that f(x)=6x^2-x-1.

Question

a)Find interval of x when k=3 and f(x) >= 0.

b) Let y=log₃ k. Show that range of of values of k is {k:1/3²⁴<^k<1} if f(x)=0 has no real roots.

Answer 1

6x^2 -x +1 = 0

b^2 -4ac = 1-24=-23<0

there are no real solutions when the discriminant <0

f(x) is always >0 for all real values of x

the interval where f(x)>0 is (-infinity, +infinity)

b) log₃k = y

3^y = k

if y= f(k) =0 has no real roots, it's graph never intersects the x or k axis

graph any logarithm and it intersects the x axis at x=1, so k has to be less than 1

k<1

but for there is no lower bound, while 1/3^24 is very small, k could be even closer to 0

0<k<1 is one interval of the domain if f(k)=0 has no real roots

(0,1)U(1, infinity)

log of k to the base 3 is very similar to the graph of log to the base e of x