Consider the recurrenceT(n).T(n)={cifn≤4T(⌊pn⌋)+ difn>4a)Use the recursion tree or repeated substitution method to come up with a good guess for a boundf(n) on the recurrenceT(n).b) State and prove by induction two theorems showing(n)∈Θ(f(n))

step-by-step explanation:

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )

total of apples - take away four

6-4 = 2

answer is 2 apples

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.