# Intro to Calculus - Why are quadratic function differentiable at the vertex?

• Feb 18th 2013, 12:23 PM
reventon
Intro to Calculus - Why are quadratic function differentiable at the vertex?
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?)
__________________________________________________ __________________________________________________ __________________________________________________ ____________________

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: The bold is 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, 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 valid, then I do not see why any function is differentiable at the local max/min (ex. Quadratic Function at the vertex)?
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)
• Feb 18th 2013, 01:24 PM
jakncoke
Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?
Its not so much that for x < 0, f(x) has a negative slope and for x > 0 f(x) has positive slope, its more the fact that as you approach 0, the negative slope and positive slopes dont go to the same limit

if you take f(x) = $x^2$ then $lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = lim_{h \to 0} \frac{x^2 + h^2 + 2xh - x^2 }{h} = lim_{h \to 0} \frac{h(h + 2x)}{h} = lim_{h \to 0} h + 2x = 2x$.

No matter how you approach anypoint in the domain, this limit will always be 2x. So the slope is defined for any point in the domain even (x=0).
• Feb 18th 2013, 01:32 PM
HallsofIvy
Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?
I'm not sure I understand your question, or, perhaps, why you are asking. What you have done is correct:
The derivative of f(x), at $x= x_0$ is, by definition, $\lim_{h\to 0}\frac{f(x_0+h)- f(x_0)}{h}$.

Yes, for f(x)= |x|, $x_0= 0$, we have to consider h< 0 and h> 0 separately because f(h)= h for h> 0, f(h)= -hf for h< 0.

So, for h> 0, we have $\lim_{h\to 0^+}\frac{|h|- 0}{h}= \lim_{h\to 0^+}\frac{h}{h}= 1$ while, for h< 0,
$\lim_{h\to 0^-}\frac{|h|- 0}{h}= \lim_{h\to 0^-}\frac{-h}{h}= -1$. Those two "one-sided" limits are not the same so the "limit" itself does not exist. |x| is not differentiable at x= 0.

Now, for $f(x)= x^2$, $x_0= 0$, that becomes $\lim_{h\to 0}\frac{(0+ h)^2- 0^2}{h}= \lim_{h\to 0}\frac{h^2}{h}= \lim_{h\to 0} h= 0$
Notice that the last "cancellation", $\frac{h^2}{h}= h$ does NOT depend upon whether h is positive or negative, unlike the case with $\frac{|h|}{h}$ which 1 or negative 1 depending upon whether h is positive or negative. Here, the "one sided limits", both "from above" and "from below", are 0. So the limit itself exists and is 0.
• Feb 18th 2013, 02:09 PM
reventon
Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?
Quote:

Originally Posted by HallsofIvy
I'm not sure I understand your question, or, perhaps, why you are asking. What you have done is correct:
The derivative of f(x), at $x= x_0$ is, by definition, $\lim_{h\to 0}\frac{f(x_0+h)- f(x_0)}{h}$.

Yes, for f(x)= |x|, $x_0= 0$, we have to consider h< 0 and h> 0 separately because f(h)= h for h> 0, f(h)= -hf for h< 0.

So, for h> 0, we have $\lim_{h\to 0^+}\frac{|h|- 0}{h}= \lim_{h\to 0^+}\frac{h}{h}= 1$ while, for h< 0,
$\lim_{h\to 0^-}\frac{|h|- 0}{h}= \lim_{h\to 0^-}\frac{-h}{h}= -1$. Those two "one-sided" limits are not the same so the "limit" itself does not exist. |x| is not differentiable at x= 0.

Now, for $f(x)= x^2$, $x_0= 0$, that becomes $\lim_{h\to 0}\frac{(0+ h)^2- 0^2}{h}= \lim_{h\to 0}\frac{h^2}{h}= \lim_{h\to 0} h= 0$
Notice that the last "cancellation", $\frac{h^2}{h}= h$ does NOT depend upon whether h is positive or negative, unlike the case with $\frac{|h|}{h}$ which 1 or negative 1 depending upon whether h is positive or negative. Here, the "one sided limits", both "from above" and "from below", are 0. So the limit itself exists and is 0.

Would you mind showing me that left and right limit are equal as x->0 for f(x) = x^2?
• Feb 18th 2013, 02:21 PM
jakncoke
Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?
Quote:

Originally Posted by reventon
Would you mind showing me that left and right limit are equal as x->0 for f(x) = x^2?

as h-->0, not x-->0

We evaluate the lim as h-->0 for f(x)= $x^2$ at point x =0.

if you take f(x) = $x^2$ then $lim_{h \to 0} \frac{f(0 + h)-f(0)}{h} = lim_{h \to 0} \frac{h^2 }{h} = lim_{h \to 0} h= 0$.
• Feb 18th 2013, 02:44 PM
reventon
Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?
Quote:

Originally Posted by jakncoke
as h-->0, not x-->0

We evaluate the lim as h-->0 for f(x)= $x^2$ at point x =0.

if you take f(x) = $x^2$ then $lim_{h \to 0} \frac{f(0 + h)-f(0)}{h} = lim_{h \to 0} \frac{h^2 }{h} = lim_{h \to 0} h= 0$.

My apologies, typo, would you mind showing me the left limit h-> 0- and right limit h-> 0+ and that they are equal for the function f(x) = x^2
• Feb 18th 2013, 03:10 PM
jakncoke
Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?
I'm gonna denote $h^{-}$ denote $h \to 0$ approaching from the left of 0.
and $h^{+}$ denote $h \to 0$ approaching from the right of 0.

Now $lim_{h^{-} \to 0} \frac{f(0 + h^{-}) - f(0)}{h^{-}} = \frac{h^{-}*h^{-}}{h^{-}} = h^{-}$ which goes to zero

$lim_{h^{+} \to 0} \frac{f(0 + h^{+}) - f(0)}{h^{+}} = \frac{h^{+}*h^{+}}{h^{+}} = h^{+}$ which goes to zero
• Feb 18th 2013, 03:33 PM
reventon
Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?
Quote:

Originally Posted by jakncoke
I'm gonna denote $h^{-}$ denote $h \to 0$ approaching from the left of 0.
and $h^{+}$ denote $h \to 0$ approaching from the right of 0.

Now $lim_{h^{-} \to 0} \frac{f(0 + h^{-}) - f(0)}{h^{-}} = \frac{h^{-}*h^{-}}{h^{-}} = h^{-}$ which goes to zero

$lim_{h^{+} \to 0} \frac{f(0 + h^{+}) - f(0)}{h^{+}} = \frac{h^{+}*h^{+}}{h^{+}} = h^{+}$ which goes to zero

So as long as the left limit and the right limit are approaching the same value, they are considered equal and a limit at 0 (in this case) would exist?
Because when I was doing this proof, I did not understand how 0+ would actually EQUAL to 0-, since they are only APPROACHING 0
• Feb 18th 2013, 03:38 PM
jakncoke
Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?
Quote:

Originally Posted by reventon
So as long as the left limit and the right limit are approaching the same value, they are considered equal and a limit at 0 (in this case) would exist?

Yes if the left and right limits approach the same value value, then f'(0) exists.

Quote:

Because when I was doing this proof, I did not understand how 0+ would actually EQUAL to 0-, since they are only APPROACHING 0
I'm assuming $0^{+}$ means a sequence approaching 0 from the right and $0^{-}$ a sequence approaching 0 from the left.
Absolutely $0^{+}$ need not equal $0^{-}$

but their limits have to be equal so,
$Lim_{h^{-} \to 0} h^{-} = Lim_{h^{+} \to 0} h^{+}$
• Feb 19th 2013, 02:22 PM
HallsofIvy
Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?
Quote:

Originally Posted by reventon
So as long as the left limit and the right limit are approaching the same value, they are considered equal and a limit at 0 (in this case) would exist?
Because when I was doing this proof, I did not understand how 0+ would actually EQUAL to 0-, since they are only APPROACHING 0

Limits do no "approach" anything- limits are numbers. In $\lim_{h\to 0^-} \frac{h^2- 0}{h}$, h is "approaching 0", and, because, for h< 0, $h^2/h= h$, the limit is equal to (not "approaching") 0. Similarly for $lim_{h\to 0^+}\frac{h^2- 0}{h}$- h "approaches" 0 but the limit is equal to 0.
• Feb 19th 2013, 08:17 PM
reventon
Re: Intro to Calculus - Why are quadratic function differentiable at the vertex?
Understood. Thank You very much jakncoke and HallsofIvy! I appreciated it.