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

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)

the weight of the object inside the box!