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Find the gradient of the tangent

  • Thread starter arvins9
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  • #1
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Homework Statement




For every x>-4 where x[tex]\in[/tex] [tex]\Re[/tex] applies

sinx+x[tex]\leq[/tex]f(x)[tex]\leq[/tex]8[tex]\sqrt{x+4}[/tex]-16

Find the gradient of the tangent to the curve of f at x[tex]_{0}[/tex]=0

Please help me I am trying to solve this exercise for more than two hours!
I'm desperate.
 

Answers and Replies

  • #2
i think the functions is g(x) <=f(x)<=h(x) has
slope [g(x) at x=0 ] = 2 and
slope [h(x) at x=0 ] = -2
however i think the function is too many in between so the question is not relevant [ i think].
 
  • #3
HallsofIvy
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i think the functions is g(x) <=f(x)<=h(x) has
slope [g(x) at x=0 ] = 2 and
slope [h(x) at x=0 ] = -2
No, [itex]h(x)= 8\sqrt{x+ 4}- 16= 8(x+4)^{1/2}- 16[/itex]
so [itex]h'(x)= 4(x+ 4)^{-1/2}[/itex] and h'(0)= 4/2= 2, not -2.

however i think the function is too many in between so the question is not relevant [ i think].
 
  • #4
No, [itex]h(x)= 8\sqrt{x+ 4}- 16= 8(x+4)^{1/2}- 16[/itex]
so [itex]h'(x)= 4(x+ 4)^{-1/2}[/itex] and h'(0)= 4/2= 2, not -2.
i think you should recheck your answer,please see h'(0) = -2:smile:
 

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