Let f(x)= {x+2 if x3 Which of the following statements are true about f? I. lim f(x) as x approaches 3 exists II. f is continuous at x=3 III. f is differentiable at x=3 the answer is I and II only, but how would you go about solving it to get the answer?

Question

Let f(x)= {x+2   if x<3
               {4x-7    if x>3

Which of the following statements are true about f?

I. lim f(x) as x approaches 3 exists
II. f is continuous at x=3
III. f is differentiable at x=3

the answer is I and II only, but how would you go about solving it to get the answer?

jigglypuff314 · Answer

you could just plug in 3 for both equations and see that they intersect at x = 3
to do it limit wise... solve for the limits of
$$\lim_{x \rightarrow 3^{-}} (x + 2)$$ and
$$\lim_{x \rightarrow 3^{+}} (4x - 7)$$
3^- part means as the limit as x is approaching 3 from the left
       which was taken from the x < 3 part

anonymous · Answer

x= 3 thn x+2 = 5
                   4x-7= 12-7=5
so f is continues as both connects in x=3 
limit f(x) exists as both coming from negative and positive sides of 3 equals to 5
its not differentiable as there are two equations and takes a sharp turn at x=3

anonymous · Answer

@harindu how do you know it takes a sharp turn at x=3 without using a calculator?

08surya · Answer

for continuity left hand limit=right hand limit=value of the function at that point

we will get left hand limit from x<3........so LHL=3+2=5
we will get right hand limit from x>3.....so RHL=4*3-7=5
and we will get value of function by puting x=3 in any expression
value of function=3+2=5
so limit exist at x=3 bcause RHL=LHL
and function is continuous becoz LHL=RHL=value of the function=3

anonymous · Answer

there are two functions and they connect at x=3 so it wont be a curve or a straight line so it would have a sharp turn