For the function to be differentiable, its derivative has to exist everywhere, which means the derivative itself must be continuous. Differentiating gives

The question mark is a placeholder, and if the derivative is to be continuous, then the question mark will have the same value as the limit as

from either side.


So the derivative will be continuous as long as

For the function to be differentiable everywhere, we need to require that

is itself continuous, which means the following limits should be the same:



So, the function should be

with derivative
Answer:
y intercept = 3
--- slope
Step-by-step explanation:
Given
The attached graph
Required
Determine the slope and the y intercept
The
is at the point where
.
From the graph:

Hence,
y intercept = 3
Next, we calculate the slope using:

Where:


So, we have:




m₁ = mass hanged initially = 0.2 kg
x₁ = initial stretch in the spring = 8 cm = 0.08 m
k = spring constant of the spring
the weight of the mass hanged is balanced by the spring force by the spring hence
Spring force = weight of mass hanged
k x₁ = m₁ g
k (0.08) = (0.2) (9.8)
k = 24.5 N/m
m₂ = mass hanged later = 0.5 kg
x₂ = final stretch in the spring = ?
k = spring constant of the spring = 24.5 N/m
the weight of the mass hanged is balanced by the spring force by the spring hence
Spring force = weight of mass hanged
k x₂ = m₂ g
(24.5) x₂ = (0.5) (9.8)
x₂ = 0.2 m = 20 cm
9514 1404 393
Answer:
∠RTU≅∠TQS
Step-by-step explanation:
To show TU ║ QS, the angles made by transversal RQ with each of those segments must be shown to be congruent. That is, it must be true that ...
∠RTU≅∠TQS
Which is the least of what!?