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 answer is C.) 4 1/4 and 5 1/2
Answer:
A. -2sin(x)sin(0.4x)
Step-by-step explanation:
At x=0, the given expression evaluates to zero.
For x near zero, the given expression is slightly negative. You know this because cos(1.4x) decreases faster than cos(0.6x), so their difference will be negative.
Of the answer choices, you can eliminate C and D because they are not zero at x=0. You can eliminate choice B, because it is positive for values of x near zero. The only viable choice is A:
-2sin(x)sin(0.4x)
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Of course, a graphing calculator can help you figure this out. See the attachment. The given expression is plotted using a solid red line; the answer choice A is plotted using blue dots.
Answer:
x=114
Step-by-step explanation:
-11=x/-6+8
-19=x/-6
x=114
21x³ - 18x²y + 24xy²
Simplify!
7x³ - 6x²y + 8xy²
