If x=-1\3 is a zero of a polynomial p(x)=27xcube -axsquare -x+3 then find the value of a
1 answer:
Answer:
a = 21
Step-by-step explanation:
If x = h is a zero of a polynomial then f(h) = 0, thus
p(- ) = 27( -
)³ - a(-
) - (-
) + 3 = 0, that is
27 (- ) -
a +
+ 3 = 0
- 1 - a +
= 0 ( multiply through by 9 to clear the fractions )
- 9 - a + 30 = 0
- a + 21 = 0 ( subtract 21 from both sides )
- a = - 21 ( multiply both sides by - 1 )
a = 21
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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.