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Romashka-Z-Leto [24]
3 years ago
12

Determine whether Rolle's Theorem can be applied to f on the closed interval

Mathematics
2 answers:
notsponge [240]3 years ago
8 0

Rolle's theorem is applicable if f(a)=f(b) and $f$ is differentiable in $(a,b)$

since it's polynomial function, it's always continuous and differentiable..

and you can easily check that $f(0)=f(-3)=0$

so it is applicable.

now, $f'(x)=-2x+3=0 \implies x=\frac32$

there is only once value (as you can imagine, the graph will be downward parabola)

Ierofanga [76]3 years ago
6 0

Answer:

Yes, Rolle's theorem can be applied

There is only one value of c such that f'(c) = 0, and this is c = 1.5 (or 3/2 in fraction form)

Step-by-step explanation:

Yes, Rolle's theorem can be applied on this function because the function is continuous in the closed interval (it is a polynomial function) and differentiable  in the open interval, and f(a) = f(b) given that:

f(0)=-0^2+3\,(0)=0\\f(3)=-3^2+3\,(3)=-9+9=0

Then there must be a c in the open interval for which f'(c) =0

In order to find "c", we derive the function and evaluate it at "c", making the derivative equal zero, to solve for c:

f(x)=-x^2+3\,x\\f'(x)=-2\,x+3\\f'(c)=-2\,c+3\\0=-2\,c+3\\2\,c=3\\c=\frac{3}{2} =1.5

There is a unique answer for c, and that is c = 1.5

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6 0
2 years ago
A number line going from negative 3 to positive 5. a closed circle is at negative 1.5. everything to the left of the circle is s
algol13

The inequalities which matches the graph are: x ≥ ₋1.5 and ₋1.5 ≤ x

Given, a number line is moving from ₋3 to ₊5 .

Next a mark is made at ₋1.5 and everything to its left is shaded which means not visible.

When we mark the point and shade the left part of it then we can start applying the inequality expressions.

And from that we can match the applicable inequalities while observing the graph.

  • For the first inequality ₋1.5 ≥ x.Here,x value ranges from ₋1.5 to ₊5, hence we take this as an inequality expression.
  • Next, if we consider x ≤ ₋1.5, then here x value will range from ₋1.5 to ₋3. where the region is shaded. Hence this expression doesn't satisfy the graph.
  • the next expression is ₋1.5 ≤ x. here the value will again range in the shaded area so it is not applicable.
  • ₋1.5 ≥ x, here the values will satisfy the graph.
  • remaining inequality expressions does not support the graph.

Therefore the only inequalities the graph represents is  x ≥ ₋1.5 and ₋1.5 ≤ x

Learn more about "Linear Inequalities" here-

brainly.com/question/371134

#SPJ10

5 0
2 years ago
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