Question

Suppose f(x) is a differentiable function such that f(2)=3 and

f'(x) is less than or equal to 4 for all x in the interval
[0,5]. Which statement below is **true** about the
function f(x)?

The Mean Value Theorem implies that f(4)=11. |
||

The Mean Value Theorem implies that f(5)=15. |
||

None of the other statements is correct. |
||

The Intermediate Value Theorem guarantees that there exists a root of the function f(x) between 0 and 5. |
||

The Intermediate Value Theorem implies that f(5) is less than or equal to 15. |

Answer #1

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4. Given the function f(x)=x^5+x-1, which of the following is
true?
The Intermediate Value Theorem implies that f'(x)=1 at some
point in the interval (0,1).
The Mean Value Theorem implies that f(x) has a root in the
interval (0,1).
The Mean Value Theorem implies that there is a horizontal
tangent line to the graph of f(x) at some point in the interval
(0,1).
The Intermediate Value Theorem does not apply to f(x) on the
interval [0,1].
The Intermediate Value Theorem...

suppose f is a differentiable function on interval (a,b) with f'(x)
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Suppose f(x)=x6+3x+1f(x)=x6+3x+1. In this problem, we will show
that ff has exactly one root (or zero) in the interval
[−4,−1][−4,−1].
(a) First, we show that f has a root in the interval
(−4,−1)(−4,−1). Since f is a SELECT ONE!!!! (continuous)
(differentiable) (polynomial) function on the
interval [−4,−1] and f(−4)= ____?!!!!!!!
the graph of y=f(x)y must cross the xx-axis at some point in the
interval (−4,−1) by the SELECT ONE!!!!!! (intermediate value
theorem) (mean value theorem) (squeeze theorem) (Rolle's theorem)
.Thus, ff...

If f is a differentiable function such that f ' ( x ) ≥ 8 , find
the largest value of M so that f ( 4 ) ≥ M whenever f ( 2 ) = 3

Does the function satisfy the hypotheses of the Mean Value
Theorem on the given interval?
f(x) = x3 + x − 9, [0, 2]
Yes, f is continuous on [0, 2] and differentiable on
(0, 2) since polynomials are continuous and differentiable on .No,
f is not continuous on [0, 2]. No,
f is continuous on [0, 2] but not differentiable on (0,
2).Yes, it does not matter if f is continuous or
differentiable; every function satisfies the Mean Value
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Does the function satisfy the hypotheses of the Mean Value
Theorem on the given interval? f(x) = 3x2 + 3x + 6, [−1, 1]
No, f is continuous on [−1, 1] but not differentiable
on (−1, 1).
There is not enough information to verify if this function
satisfies the Mean Value Theorem.
Yes, it does not matter if f is continuous or
differentiable; every function satisfies the Mean Value
Theorem.
No, f is not continuous on [−1, 1].Yes, f is...

To illustrate the Mean Value Theorem with a specific function,
let's consider f(x) = x^3 − x, a = 0, b = 5. Since f is a
polynomial, it is continuous and differentiable for all x, so it is
certainly continuous on [0, 5] and differentiable on (0, 5).
Therefore, by the Mean Value Theorem, there is a number c in (0, 5)
such that
f(5) − f(0) = f '(c)(5 − 0).
Now f(5) = ______ , f(0) =...

1aDoes the function satisfy the hypotheses of the Mean Value
Theorem on the given interval?
f(x) = 4x2 + 3x + 1, [−1,
1]
a.No, f is continuous on [−1, 1] but not differentiable
on (−1, 1).
b.Yes, it does not matter if f is continuous or
differentiable; every function satisfies the Mean Value
Theorem.
c.Yes, f is continuous on [−1, 1] and differentiable on
(−1, 1) since polynomials are continuous and differentiable on
.
d.No, f is not continuous on...

The function f ( x ) = 3 x ^3 + 5 x + 12 satisfies
the hypotheses of the Mean Value Theorem on the interval [0, 2].
Find all value(s) c that satisfy the conclusion of the theorem.

If f is a differentiable function such that f′(x) = (x^2−
16)*g(x), where g(x)>0 for all x, at which value(s) of x does f
have a local maximum?
1. At both x=-4,4
2. Only at x=-16
3. Only at x=4
4. At both x=-16,16
5. Only at x=-4
6. Only at x=16

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