Question

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​?

Answer 1

Answer:

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