This makes sense if you’re looking at how the tangents change from left to right on the curves. If you look at the tangents from right to left it breaks down. On the concave down shape, from right to left, the tangent is decreasing to 0 then getting steeper but in a positive way.
Actually, there are TWO cases where the Second Derivative Test is inconclusive: (1) f"(c) = 0 (you have that one), and (2) f"(c) undefined (you missed that one). It IS possible for f'(c) to be DEFINED but f"(c) to be UNDEFINED. For example, consider the function f(x) = x^(4/3). Then f'(x) is essentially x^(1/3) which is 0 only for x = 0, so x = 0 is a critical number (the only one, since f'(x) is never undefined). Then, f"(x) is essentially x^(-2/3), which is UNDEFINED for x = 0, so the Second Derivative Test is INCONCLUSIVE for the only critical number for f, x = 0. However, if we use the First Derivative Test, we find that f'(x) is negative for x to the left of 0 and positive for x to the right of 0, so clearly (0,0) is a local MINIMUM point.
Let’s for example use x^2 f’x=2x so the rate of change of our function is 2x. The second derivative would be 2, which would be the rate of change of our first derivative which is constantly changing by a factor of 2.
@@jimmybob609 yes. Your comment helped out a lot dy/dx is y/x also known as rate Example d/t = S (distance travelled per unit time) is rate. Also gives you a slope at any given point (d/t)/t = second derivative which tells you what happens as the rate changes over time eg: speed/t = acceleration. is the change of speed
This makes sense if you’re looking at how the tangents change from left to right on the curves. If you look at the tangents from right to left it breaks down. On the concave down shape, from right to left, the tangent is decreasing to 0 then getting steeper but in a positive way.
Actually, there are TWO cases where the Second Derivative Test is inconclusive: (1) f"(c) = 0 (you have that one), and (2) f"(c) undefined (you missed that one). It IS possible for f'(c) to be DEFINED but f"(c) to be UNDEFINED. For example, consider the function f(x) = x^(4/3). Then f'(x) is essentially x^(1/3) which is 0 only for x = 0, so x = 0 is a critical number (the only one, since f'(x) is never undefined). Then, f"(x) is essentially x^(-2/3), which is UNDEFINED for x = 0, so the Second Derivative Test is INCONCLUSIVE for the only critical number for f, x = 0. However, if we use the First Derivative Test, we find that f'(x) is negative for x to the left of 0 and positive for x to the right of 0, so clearly (0,0) is a local MINIMUM point.
Bless Khan Academy during corona virus pandemic 🙏🙏🙏
thank you very Much, so helpful
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Umair facts it looks really nice
Lowkey makes you more interested and makes it less monotonous
Sir, thanks for _existing_
Why is 2nd derivative of functions -ve for max and +ve for min
What is the higher operation needed to determine the concavity of a function with a critical point of 0?
Noone knows answer
How u set pen for writing
this might be obvious but how did you get -4 for the second derivative?
It's a hypothetical.
its just an example
do sal do this with a mouse?
no...
wait. Isn't the second diff of a quadratic a point?
Let’s for example use x^2 f’x=2x so the rate of change of our function is 2x. The second derivative would be 2, which would be the rate of change of our first derivative which is constantly changing by a factor of 2.
@@jimmybob609
yes. Your comment helped out a lot
dy/dx is y/x also known as rate
Example
d/t = S (distance travelled per unit time) is rate. Also gives you a slope at any given point
(d/t)/t = second derivative which tells you what happens as the rate changes over time
eg: speed/t = acceleration. is the change of speed
@@jimmybob609
wait. clarify this for me
the second derivative determins how pointy is a function for x^3? and x^2 right?