What is the proof for f(1.1)>-0.1?

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SUMMARY

The discussion focuses on proving that for a continuous and differentiable function f(x) of order 2, where f''(x) > 0, the condition f(1.1) > -0.1 holds true given the tangent line at x=1 is y=-x+1. The function can be expressed as f(x) = ax² + bx + c, with parameters a > 0, f(1) = 0, and f'(1) = -1. By substituting these values, it is established that f(1.1) = 0.01a - 0.1, which confirms that f(1.1) is indeed greater than -0.1.

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Yankel
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Hello all,

I am not sure how to approach this question:

Let f(x) be a continuous and differentiable function of order 2. Let f''(x) >0 for all values of x. The tangent line to the function at x=1 is y=-x+1. Show that f(1.1)>-0.1.

Thanks!
 
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I would begin with:

$$f(x)=ax^2+bx+c$$

You are given:

$$a>0$$

$$f(1)=0$$

$$f'(1)=-1$$

Use these data to write $b$ and $c$ in terms of $a$. You should then be able to demonstrate that:

$$f(1.1)=0.01a-0.1>-0.1$$
 

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