Good day everybody!! (Hi)

I am new to the forum and I just started calculus for a month or two. We are currently learning limits and derivatives and I came across this problem (or perhaps a misunderstanding on my part (Itwasntme))

Why are quadratic functions (or any other functions with a local max/min) differentiable at the vertex (or at their local max/min)?

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Before I start, there are a couple things that I would like to note before going further, I was taught that:

A limit only exists when:

1) The left limit exists

2) The right limit exists

3) The left and right limits are equal

If any of these conditions are not met, a limit does not exist (DNE)

In addition, at least from my instructor, I was told that infinity is regarded as a "DNE". A limit is a real number.

A function is differentiable at a if f'(a) exist (The tangent exist at x=a)

All differentiable function must be continuous.

NOT all continuous functions are differentiable.

A function is not differentiable at points where:

1) A tangent does not exist

2) The function is discontinuous

3) The left and right limits of the slope of the secant are not equal (<-- The wording of this one is what bothers me the most, does it simply mean the left limit and right limit are not equal as x -> point of tangency?)

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Okay, with that being said, here is my problem.

For the absolute value function, ex. f(x) = abs(x), it is known that at x = 0, the function is not differentiable as "the tangent does not exist because the left and right limits of the slope of the secant does not equal" We did a proof in class:

(Note: Theboldis to emphasize the numerator for easier reading.

Left Limit

lim(h->0-)f(x+h)-f(x)/ h

lim(h->0-)-(x+h)-(-x)/ h <- by multiplying -1, the absolute value sign (| |) is replaced

lim(h->0-)-x-h+x/ h

lim(h->0-)-h/ h

lim(h->0-)-1

-1

Right Limit

lim(h->0+)f(x+h)-f(x)/ h

lim(h->0+)(x+h)-(x)/ h <- given x = 0 and h > 0 (Positive value), absolute value is redundant here

lim(h->0+)x+h-x/ h

lim(h->0+)h/ h

lim(h->0+)1

1

Since LL /= RL as x->0

The function is not differentiable at x = 0

From this proof,If the above statement is valid, then I do not see why any function is differentiable at the local max/min (ex. Quadratic Function at the vertex)?I am unsure if I can conclude that at x = 0, the function is not differentiable because for the interval x < 0, f(x) has a negative slope and for the interval x > 0, f(x) has a positive slope.

If the above statement is invalid, then why is the absolute value function f(x) = abs(x) indifferentiable at x = 0?

Thanks in advance and your time and effort is much appreciated!! (Clapping)