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

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To prove that f(1.1) > -0.1 for the continuous and differentiable function f(x), we start with the given conditions: f''(x) > 0 indicates that f(x) is concave up, and the tangent line at x=1 is y = -x + 1, leading to f(1) = 0 and f'(1) = -1. By expressing f(x) in the form f(x) = ax^2 + bx + c, and using the conditions to derive b and c in terms of a, we find that f(1.1) simplifies to 0.01a - 0.1. Since a > 0, it follows that 0.01a is positive, confirming that f(1.1) is indeed greater than -0.1. Thus, the proof is established.
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:

$$f(1.1)=0.01a-0.1>-0.1$$
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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