Let f be the following piecewise-defined function. f(x)equals left brace start 2 by 1 matrix 1st row 1st column x squared plus 5 2nd row 1st column 3 x plus 5 endmatrix for xless than or equals3 for xgreater than3 (a) is f continuous at xequals3? (b) is f differentiable at xequals3?

the function is continuous but not differentiable  in x=3

Step-by-step explanation:

for f(x) defined as

x²+5 for x≤3

3*x+5 for x>3

then for f to be continuous then

when x→a , lim f(x) = f(a)

thus

from the left

when x→3⁻ , lim f(x) = lim x²+5 = 3²+5 = 14

from the right

when x→3⁺ , lim f(x) = lim 3*x+5 = 3*3+5 = 14

and f(3) = 3²+5 = 14

since both limits converge and are equal to f(3), then the function is continuous in 3

Nevertheless, the function is not differentiable in 3 since the curve is not smooth in f(3) ( it has an sharp change in f(3) ) , then the function is not differentiable  in x=3

to prove it , when x=3 and

when h→0⁻ , f'(x=3)= lim [f(x+h)-f(x)] /h  = lim [(x+h)²+5 - (x² + 5)]/h = lim [2*x*h+h²] /h = lim [2*x+h] = 2*3+0 = 6

when h→0⁺ , f'(x=3)= lim [f(x+h)-f(x)] /h  = lim [3*(x+h)+5 - (3*x + 5)]/h = lim [3*h] /h = lim 3 = 3

since the limits do not converge →the limit does not exist → the derivative does not exist → the function is not differentiable in x=3

i think it would be 43 movies that would have to be watched.

step-by-step explanation:

think about it. for the second plan, every month is 8 dollars. 8 x 12= 96. so thats the number for the second plan. now for the first plan, its annual meaning you have to pay 10 dollars every year. so now you have to get an equivalent amount of 86 since 96-10=86. 86 divided by 2 since every movie costs 2$ then you would have to watch 43 movies a year to get the same as the second plan. i really hoped this . have a wonderful day as well!

its d. the first coordintaes are 10,32

step-by-step explanation: