How do you determine all values of c that satisfy the mean value theorem on the interval [-1,1] for #f(x) = 3x^5+5x^3+15x #?

Answer 1

To determine all values of ( c ) that satisfy the Mean Value Theorem on the interval ([-1, 1]) for ( f(x) = 3x^5 + 5x^3 + 15x ), you need to follow these steps:

  1. Find the derivative of ( f(x) ), denoted as ( f'(x) ).
  2. Calculate the average rate of change of ( f(x) ) over the interval ([-1, 1]), which is given by: [ \frac{f(1) - f(-1)}{1 - (-1)} ]
  3. Find all values of ( c ) such that ( f'(c) ) equals the average rate of change calculated in step 2.

So, here are the steps more specifically:

  1. Find the derivative: [ f'(x) = 15x^4 + 15x^2 + 15 ]

  2. Calculate the average rate of change: [ \text{Average rate of change} = \frac{f(1) - f(-1)}{1 - (-1)} ] [ = \frac{f(1) - f(-1)}{2} ]

Now, calculate ( f(1) ) and ( f(-1) ): [ f(1) = 3(1)^5 + 5(1)^3 + 15(1) = 3 + 5 + 15 = 23 ] [ f(-1) = 3(-1)^5 + 5(-1)^3 + 15(-1) = -3 - 5 - 15 = -23 ]

Substitute these values into the formula for the average rate of change: [ \text{Average rate of change} = \frac{23 - (-23)}{2} = \frac{46}{2} = 23 ]

  1. Find ( c ) such that ( f'(c) = 23 ): [ 15c^4 + 15c^2 + 15 = 23 ]

Solve this equation to find all possible values of ( c ).

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Answer 2

There are two values #c=+-0.62#

On the interval [-1,1], the function #f(x)# is defined and continuous, and differentiable on the interval [-1,1] as it is a polynomial function. So we can apply the mean value theorem.
There is value #c ∈ [-1,1]# such that #f'(c)=(f(1)-f(-1))/(1-(-1))#
Let's determine #f(1)=3+5+15=23# And #f(-1)=-3-5-15=-23#
So #f'(c)=(23--23)/2=46/2=23#
Let's calculate #f'(x)=15x^4+15x^2+15=15(x^4+x^2+1)# Then #f'(c)=15(c^4+c^2+1)=23#
#c^4+c^2+1=23/15# #c^4+c^2-8/15=0# Rewrite as #15c^4+15c^2-8=0# let #c^2=y# #15y^2+15y-8=0# #Delta = 15^2+4*15*8=705# So #y=(-15+-sqrt705)/30=(-15+26.6)/30=11.6/30=0.39#

We take only the positive root

Then #c=+-sqrt0.39=+-0.62#
#+-0.62 ∈[-1,1]#
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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