Answer:
Yes , function is continuous in [0,2] and is differentiable (0,2) since polynomial function are continuous and differentiable
Step-by-step explanation:
We are given the Function
f(x) =
The two basic hypothesis of the mean valued theorem are
- The function should be continuous in [0,2]
- The function should be differentiable in (1,2)
upon checking the condition stated above on the given function
f(x) is continuous in the interval [0,2] as the functions is quadratic and we can conclude that from its graph
also the f(x) is differentiable in (0,2)
f'(x) = 6x - 2
Now the function satisfies both the conditions
so applying MVT
6x-2 = f(2) - f(0) / 2-0
6x-2 = 9 - 1 /2
6x-2 = 4
6x=6
x=1
so this is the tangent line for this given function.
the slope is 1 and the y intercept is-2
We are asked to find the equivalent of the expression given:
(3m⁻² n)⁻³
-----------
6mn⁻²
Perform distribution of power using power rule such as shown below:
3⁻³ m⁻²*⁻³n⁻³
-----------------
6mn⁻²
Perform product and quotient rule such as shown below:
m⁶ n²
--------
3³ *6*m*n³
Simplify,
m⁵
--------
162n
The answer is m⁵/162n.
Answer:
A = 2
B = 4
C = 1
D = 3
Step-by-step explanation:
Answer: |5x+1|-7
Step-by-step explanation:
