I am struggling with the following questions:Let g ( x ) be continuous in [ a , b ] and 0 ≤ r ≤ 1 I need to prove that there is x 0 ∈ [ a , b ] that for him exists: g ( x 0 ) = r g ( a ) + ( 1 − r ) g ( b )I know that for r = 0 and r = 1. I am getting immediate proof of that(it is the easy part). Where I am struggling is finding and understanding the values of r I need to use. I've tried defining a helping function g ( x ) = r g ( a ) + ( 1 − r ) g ( b ) − g ( x 0 ) but no success on that end. I need to show that for 2 different values in the range of 0 &lt; r &lt; 1 one time I am getting a positive result and for the other a negative.I am not asking for the solution, asking for guidance on how it can be done.

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I am struggling with the following questions:Let   g  (  x  ) be continuous in   [  a  ,  b  ] and   0  ≤  r  ≤  1 I need to prove that there is       x    0    ∈  [  a  ,  b  ] that for him exists:   g  (      x    0    )  =  r  g  (  a  )  +  (  1  −  r  )  g  (  b  )I know that for   r  =  0 and   r  =  1. I am getting immediate proof of that(it is the easy part). Where I am struggling is finding and understanding the values of   r I need to use. I&#039;ve tried defining a helping function   g  (  x  )  =  r  g  (  a  )  +  (  1  −  r  )  g  (  b  )  −  g  (      x    0    ) but no success on that end. I need to show that for 2 different values in the range of   0  &amp;lt;  r  &amp;lt;  1 one time I am getting a positive result and for the other a negative.I am not asking for the solution, asking for guidance on how it can be done.

LitikoIDu6 · Accepted Answer

Consider   f  (  r  )  =  r  g  (  a  )  +  (  1  −  r  )  g  (  b  )  . Note that for   r  ∈  [  0  ,  1  ]  ,  f  (  r  ) is necessarily between   g  (  a  ) and   g  (  b  ). Then because   f  (  r  ) is between   g  (  a  ) and   g  (  b  ), by the Intermediate Value Theorem applied to   g, we know that for some       x    0    ∈  [  a  ,  b  ]  ,  g  (      x    0    )  =  f  (  r  )  =  r  g  (  a  )  +  (  1  −  r  )  g  (  b  ), as requiblack.

asigurato7 · Accepted Answer

Key Realization: In general for any interval   [  x  ,  y  ], any point   z  ∈  [  x  ,  y  ] can be expressed by  z  =  (  1  −  λ  )  x  +  λ  yfor some   λ  ∈  [  0  ,  1  ]. (Think of   λ as measuring a percentage distance traveled from   x to   y.)Sketch:1. With the above, what can you conclude about where   r  g  (  a  )  +  (  1  −  r  )  g  (  b  ) lives, regardless of the value of   r  ∈  [  0  ,  1  ]?2. Since   g is continuous on   [  a  ,  b  ], alongside the realization from the previous line, the intermediate value theorem applies and therefore what follows?

Let g(x) be continuous in [a,b] and 0<=r<=1 I need to prove that there is x_0 in [a,b] that for him exists: g(x_0)=rg(a)+(1−r)g(b)

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New Questions in Calculus 2