Question

Medal!!
Use the graph of f(x) = |x(x^2 − 1)| to find how many numbers in the interval [0.5, 0.75] satisfy the conclusion of the Mean Value Theorem.

1
2
4
None

Answer 1

@Michele_Laino

Answer 2

its answer 1

Answer 3

how do u know?

Answer 4

Did you mean |x(x^2-1)| ?

f(x) is continuous for all real x
f(x) is differentiable for all x ≠ −1, 0, or 1

f(x) is continuous on interval [1/2, 3/4]
f(x) is differentiable on interval (1/2, 3/4)
On the interval [1/2, 3/4]
f(x) = −x(x^2−1) = −x^3+x
f'(x) = −3x^2 + 1

(f(3/4)−f(1/2))/(3/4−1/2) = (21/64−3/8)/(1/4) = −3/16

Now we find all values of x on interval (1/2, 3/4) so that
f'(x) = −3/16
−3x^2 + 1 = −3/16
3x^2 = 19/16
x^2 = 19/48
x = √(19/48) = 0.62915287 ---> ok
[We ignore negative root, since it is not in the given interval]

so yea its answer 1 @trisarahtops

Answer 5

@Trisarahtops

Answer 6

is that right @Michele_Laino

Answer 7

sure double checking is good

Answer 8

If I explicit the absolute value, I can write this:
\[f\left( x \right) = \begin{array}{*{20}{c}}
{x\left( {{x^2} - 1} \right),\quad x \in \left( { - 1,0} \right) \cup \left( {1, + \infty } \right)} \\
{x\left( {1 - {x^2}} \right),\quad x \in \left( { - \infty , - 1} \right) \cup \left( {0,1} \right)}
\end{array}\]

Answer 9

so 2?

Answer 10

inside the interval \((1/2,3/4)\) we have two definitions for \(f(x)\)

Answer 11

so we have to do the computations separately, each computation for each subinterval

Answer 12

here is the first computation:
\[\frac{{f\left( 1 \right) - f\left( {1/2} \right)}}{{1 - \left( {1/2} \right)}} = 1 - 3{c^2}\]
where
I have used:
\[f\left( x \right) = x\left( {1 - {x^2}} \right)\]

Answer 13

f(1)-f(3/4) / 1-(3/4)

Answer 14

after a simple computation I get this:
\[c = \sqrt {\frac{{19}}{{48}}} \simeq 0.63 < 1\]
as we can see such value of \(c\) is greater than 0.5, and it is less than 1, so it is an acceptable value

Answer 15

next we have to do this computation:
\[\frac{{f\left( {3/4} \right) - f\left( 1 \right)}}{{\left( {3/4} \right) - 1}} = 1 - 3{c^2}\]
where:
\[f\left( x \right) = x\left( {1 - {x^2}} \right)\]

Answer 16

so 1 is the answer

Answer 17

please wait, I'm doing the remaining computation
after a simple computation, I get:

Answer 18

ok

Answer 19

I got this:
\[c = \sqrt {\frac{{149}}{{128 \cdot 3}}} \simeq 0.623 < 3/4\]
being such value less than \(3/4\), it can not be considered an acceptable value, so I confirm the answer of @nikeboi101

Answer 20

Thank you for clearing that up :D

Answer 21

:)

Answer 22

c:

Answer 23

you too ;)