Find the gradient of the tangent

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Homework Help Overview

The problem involves determining the gradient of the tangent to the function f(x) within the constraints of the inequalities sin(x) + x ≤ f(x) ≤ 8√(x + 4) - 16 for x > -4, specifically at x₀ = 0. The original poster expresses frustration after spending considerable time on the exercise.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the slopes of the bounding functions g(x) and h(x) at x = 0, with some suggesting that the question may be overly complex or irrelevant due to the number of functions involved. There is also a focus on verifying the derivatives of h(x) and the implications for the gradient.

Discussion Status

The discussion is ongoing, with participants questioning each other's calculations regarding the slopes of the bounding functions. There is a lack of consensus on the correct slope values, and some participants are encouraging re-evaluation of the derivatives.

Contextual Notes

Participants are navigating through potential misinterpretations of the function's behavior and the implications of the inequalities provided in the problem statement.

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Homework Statement

For every x>-4 where x\in \Re applies

sinx+x\leqf(x)\leq8\sqrt{x+4}-16

Find the gradient of the tangent to the curve of f at x_{0}=0

Please help me I am trying to solve this exercise for more than two hours!
I'm desperate.
 
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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
however i think the function is too many in between so the question is not relevant [ i think].
 
nik21bigbang said:
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, h(x)= 8\sqrt{x+ 4}- 16= 8(x+4)^{1/2}- 16
so h&#039;(x)= 4(x+ 4)^{-1/2} 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].
 
HallsofIvy said:
No, h(x)= 8\sqrt{x+ 4}- 16= 8(x+4)^{1/2}- 16
so h&#039;(x)= 4(x+ 4)^{-1/2} 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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