On Differentiable Functions (Khan Academy)
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I am learning Khan Academy Calculus, specifically how to tell when the graph of a function is differentiable or not. Khan Academy tells me that a point on the graph of is not differentiable when: 
But I don't get why 2. and 3. are true. For 2, say we have a vertical tangent line for function 1/x^2 
Would it not be continuous here? And therefore, wouldn't there be at least the chance that it's differentiable? Can someone please explain to me why a function is never differentiable at the vertical asymptote?
For 3, why would a function not be differentiable when it has a sharp turn? I don't get this either. Can someone please explain?
calculus derivatives continuity
asked Sep 2 at 10:05
Ethan Chan
702323
add a comment |Â
up vote
2
down vote
favorite
I am learning Khan Academy Calculus, specifically how to tell when the graph of a function is differentiable or not. Khan Academy tells me that a point on the graph of is not differentiable when:
But I don't get why 2. and 3. are true. For 2, say we have a vertical tangent line for function 1/x^2
Would it not be continuous here? And therefore, wouldn't there be at least the chance that it's differentiable? Can someone please explain to me why a function is never differentiable at the vertical asymptote?
For 3, why would a function not be differentiable when it has a sharp turn? I don't get this either. Can someone please explain?
calculus derivatives continuity
asked Sep 2 at 10:05
Ethan Chan
702323
For 2. Do you the slope of a vertical line? For 3. $lim_a^- f'(x)neqlim_a^+ f'(x)$ which is the basic condition for differentiability.
– prog_SAHIL
Sep 2 at 10:11
At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$†isn’t even a statement at all, so it couldn’t be either wrong or false.
– Michael Hoppe
Sep 2 at 10:18
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am learning Khan Academy Calculus, specifically how to tell when the graph of a function is differentiable or not. Khan Academy tells me that a point on the graph of is not differentiable when:
But I don't get why 2. and 3. are true. For 2, say we have a vertical tangent line for function 1/x^2
Would it not be continuous here? And therefore, wouldn't there be at least the chance that it's differentiable? Can someone please explain to me why a function is never differentiable at the vertical asymptote?
For 3, why would a function not be differentiable when it has a sharp turn? I don't get this either. Can someone please explain?
calculus derivatives continuity
asked Sep 2 at 10:05
Ethan Chan
702323
I am learning Khan Academy Calculus, specifically how to tell when the graph of a function is differentiable or not. Khan Academy tells me that a point on the graph of is not differentiable when:
But I don't get why 2. and 3. are true. For 2, say we have a vertical tangent line for function 1/x^2
Would it not be continuous here? And therefore, wouldn't there be at least the chance that it's differentiable? Can someone please explain to me why a function is never differentiable at the vertical asymptote?
For 3, why would a function not be differentiable when it has a sharp turn? I don't get this either. Can someone please explain?
calculus derivatives continuity
calculus derivatives continuity
asked Sep 2 at 10:05
Ethan Chan
702323
asked Sep 2 at 10:05
Ethan Chan
702323
asked Sep 2 at 10:05
Ethan Chan
702323
asked Sep 2 at 10:05
Ethan Chan
702323
asked Sep 2 at 10:05
Ethan Chan
702323
702323
For 2. Do you the slope of a vertical line? For 3. $lim_a^- f'(x)neqlim_a^+ f'(x)$ which is the basic condition for differentiability.
– prog_SAHIL
Sep 2 at 10:11
At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$†isn’t even a statement at all, so it couldn’t be either wrong or false.
– Michael Hoppe
Sep 2 at 10:18
add a comment |Â
For 2. Do you the slope of a vertical line? For 3. $lim_a^- f'(x)neqlim_a^+ f'(x)$ which is the basic condition for differentiability.
– prog_SAHIL
Sep 2 at 10:11
At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$†isn’t even a statement at all, so it couldn’t be either wrong or false.
– Michael Hoppe
Sep 2 at 10:18
For 2. Do you the slope of a vertical line? For 3. $lim_a^- f'(x)neqlim_a^+ f'(x)$ which is the basic condition for differentiability.
– prog_SAHIL
Sep 2 at 10:11
For 2. Do you the slope of a vertical line? For 3. $lim_a^- f'(x)neqlim_a^+ f'(x)$ which is the basic condition for differentiability.
– prog_SAHIL
Sep 2 at 10:11
At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$†isn’t even a statement at all, so it couldn’t be either wrong or false.
– Michael Hoppe
Sep 2 at 10:18
At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$†isn’t even a statement at all, so it couldn’t be either wrong or false.
– Michael Hoppe
Sep 2 at 10:18
add a comment |Â
2 Answers
2
active
oldest
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up vote
3
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accepted
For your given function $f(x)=frac1x^2$ this is not differentiable at 0 since the function isn't even continuous there since at 0 it is undefined. It approaches infinity from either side leading to a infinite gradient on either side but AT 0 it does not exist.
For (3) I would think a better way of saying this is a 'kink' for example take $f(x)=|x|$. This is not differentiable at 0 since from 0+ we have gradient 1, and -1 from 0-. Or you can think about drawing lots of tangents at the kink, they all look like tangents but are a all different lines so it is not differentiable.
answered Sep 2 at 10:11
Fundamentals
835
Oh right. Okay last question. How about for 3? Why would a function not be differentiable when there's a "sharp turn"?
– Ethan Chan
Sep 2 at 10:13
This is explained above, a very sharp turn means i can find lots of tangent lines so it would not be differentiable.
– Fundamentals
Sep 2 at 10:15
1
Okay thanks. I'll accept this answer in a minute.
– Ethan Chan
Sep 2 at 10:18
Also, we want the limit to exist that is f(x)-f(0)/x as x goes to 0. That is why f(0) existing is necessary.
– Fundamentals
Sep 2 at 10:23
add a comment |Â
up vote
0
down vote
At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$.†isn’t even a statement at all, so it can be neither wrong nor right.
answered Sep 2 at 10:20
Michael Hoppe
9,70631532
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For your given function $f(x)=frac1x^2$ this is not differentiable at 0 since the function isn't even continuous there since at 0 it is undefined. It approaches infinity from either side leading to a infinite gradient on either side but AT 0 it does not exist.
For (3) I would think a better way of saying this is a 'kink' for example take $f(x)=|x|$. This is not differentiable at 0 since from 0+ we have gradient 1, and -1 from 0-. Or you can think about drawing lots of tangents at the kink, they all look like tangents but are a all different lines so it is not differentiable.
answered Sep 2 at 10:11
Fundamentals
835
Oh right. Okay last question. How about for 3? Why would a function not be differentiable when there's a "sharp turn"?
– Ethan Chan
Sep 2 at 10:13
This is explained above, a very sharp turn means i can find lots of tangent lines so it would not be differentiable.
– Fundamentals
Sep 2 at 10:15
1
Okay thanks. I'll accept this answer in a minute.
– Ethan Chan
Sep 2 at 10:18
Also, we want the limit to exist that is f(x)-f(0)/x as x goes to 0. That is why f(0) existing is necessary.
– Fundamentals
Sep 2 at 10:23
add a comment |Â
up vote
3
down vote
accepted
For your given function $f(x)=frac1x^2$ this is not differentiable at 0 since the function isn't even continuous there since at 0 it is undefined. It approaches infinity from either side leading to a infinite gradient on either side but AT 0 it does not exist.
For (3) I would think a better way of saying this is a 'kink' for example take $f(x)=|x|$. This is not differentiable at 0 since from 0+ we have gradient 1, and -1 from 0-. Or you can think about drawing lots of tangents at the kink, they all look like tangents but are a all different lines so it is not differentiable.
answered Sep 2 at 10:11
Fundamentals
835
Oh right. Okay last question. How about for 3? Why would a function not be differentiable when there's a "sharp turn"?
– Ethan Chan
Sep 2 at 10:13
This is explained above, a very sharp turn means i can find lots of tangent lines so it would not be differentiable.
– Fundamentals
Sep 2 at 10:15
1
Okay thanks. I'll accept this answer in a minute.
– Ethan Chan
Sep 2 at 10:18
Also, we want the limit to exist that is f(x)-f(0)/x as x goes to 0. That is why f(0) existing is necessary.
– Fundamentals
Sep 2 at 10:23
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For your given function $f(x)=frac1x^2$ this is not differentiable at 0 since the function isn't even continuous there since at 0 it is undefined. It approaches infinity from either side leading to a infinite gradient on either side but AT 0 it does not exist.
For (3) I would think a better way of saying this is a 'kink' for example take $f(x)=|x|$. This is not differentiable at 0 since from 0+ we have gradient 1, and -1 from 0-. Or you can think about drawing lots of tangents at the kink, they all look like tangents but are a all different lines so it is not differentiable.
answered Sep 2 at 10:11
Fundamentals
835
For your given function $f(x)=frac1x^2$ this is not differentiable at 0 since the function isn't even continuous there since at 0 it is undefined. It approaches infinity from either side leading to a infinite gradient on either side but AT 0 it does not exist.
For (3) I would think a better way of saying this is a 'kink' for example take $f(x)=|x|$. This is not differentiable at 0 since from 0+ we have gradient 1, and -1 from 0-. Or you can think about drawing lots of tangents at the kink, they all look like tangents but are a all different lines so it is not differentiable.
answered Sep 2 at 10:11
Fundamentals
835
answered Sep 2 at 10:11
Fundamentals
835
answered Sep 2 at 10:11
Fundamentals
835
answered Sep 2 at 10:11
Fundamentals
835
835
Oh right. Okay last question. How about for 3? Why would a function not be differentiable when there's a "sharp turn"?
– Ethan Chan
Sep 2 at 10:13
This is explained above, a very sharp turn means i can find lots of tangent lines so it would not be differentiable.
– Fundamentals
Sep 2 at 10:15
1
Okay thanks. I'll accept this answer in a minute.
– Ethan Chan
Sep 2 at 10:18
Also, we want the limit to exist that is f(x)-f(0)/x as x goes to 0. That is why f(0) existing is necessary.
– Fundamentals
Sep 2 at 10:23
add a comment |Â
Oh right. Okay last question. How about for 3? Why would a function not be differentiable when there's a "sharp turn"?
– Ethan Chan
Sep 2 at 10:13
This is explained above, a very sharp turn means i can find lots of tangent lines so it would not be differentiable.
– Fundamentals
Sep 2 at 10:15
1
Okay thanks. I'll accept this answer in a minute.
– Ethan Chan
Sep 2 at 10:18
Also, we want the limit to exist that is f(x)-f(0)/x as x goes to 0. That is why f(0) existing is necessary.
– Fundamentals
Sep 2 at 10:23
Oh right. Okay last question. How about for 3? Why would a function not be differentiable when there's a "sharp turn"?
– Ethan Chan
Sep 2 at 10:13
Oh right. Okay last question. How about for 3? Why would a function not be differentiable when there's a "sharp turn"?
– Ethan Chan
Sep 2 at 10:13
This is explained above, a very sharp turn means i can find lots of tangent lines so it would not be differentiable.
– Fundamentals
Sep 2 at 10:15
This is explained above, a very sharp turn means i can find lots of tangent lines so it would not be differentiable.
– Fundamentals
Sep 2 at 10:15
1
1
Okay thanks. I'll accept this answer in a minute.
– Ethan Chan
Sep 2 at 10:18
Okay thanks. I'll accept this answer in a minute.
– Ethan Chan
Sep 2 at 10:18
Also, we want the limit to exist that is f(x)-f(0)/x as x goes to 0. That is why f(0) existing is necessary.
– Fundamentals
Sep 2 at 10:23
Also, we want the limit to exist that is f(x)-f(0)/x as x goes to 0. That is why f(0) existing is necessary.
– Fundamentals
Sep 2 at 10:23
add a comment |Â
up vote
0
down vote
At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$.†isn’t even a statement at all, so it can be neither wrong nor right.
answered Sep 2 at 10:20
Michael Hoppe
9,70631532
add a comment |Â
up vote
0
down vote
At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$.†isn’t even a statement at all, so it can be neither wrong nor right.
answered Sep 2 at 10:20
Michael Hoppe
9,70631532
add a comment |Â
up vote
0
down vote
up vote
0
down vote
At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$.†isn’t even a statement at all, so it can be neither wrong nor right.
answered Sep 2 at 10:20
Michael Hoppe
9,70631532
At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$.†isn’t even a statement at all, so it can be neither wrong nor right.
answered Sep 2 at 10:20
Michael Hoppe
9,70631532
answered Sep 2 at 10:20
Michael Hoppe
9,70631532
answered Sep 2 at 10:20
Michael Hoppe
9,70631532
answered Sep 2 at 10:20
Michael Hoppe
9,70631532
9,70631532
add a comment |Â
add a comment |Â
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For 2. Do you the slope of a vertical line? For 3. $lim_a^- f'(x)neqlim_a^+ f'(x)$ which is the basic condition for differentiability.
– prog_SAHIL
Sep 2 at 10:11
At $x=0$ the function isn’t defined, so the arrangement of number, symbol and words “$f$ is not continuous at $x=0$†isn’t even a statement at all, so it couldn’t be either wrong or false.
– Michael Hoppe
Sep 2 at 10:18