MHB Finding all points where tangent line is perpendicular

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To find points on the function f(x) = (x - √π)/(x + 1) where the tangent lines are perpendicular to the line y = -(1 + √πx + 7πe^(e^(π^110))), the derivative f'(x) is calculated as f'(x) = (1 + √π)/(x^2 + 2x + 1). The next step involves setting this derivative equal to the negative reciprocal of the slope of the given line, which is -1/(1 + √π). This leads to the equation (1 + √π)/(x^2 + 2x + 1) = -1/(1 + √π). Cross-multiplying will help simplify and solve for the x-values where the tangent lines are perpendicular.
riri
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Hello!

I've encountered a problem of find all points (x,y) on $f(x)=\frac{x-\sqrt{\pi}}{x+1}$ where there are tangent lines perpendicular to $y=-(1+\sqrt{\pi}x+7\pi e^{e^{{\pi}^{110}}})$

So I first found derivative and ended up with $f'(x)=\frac{1(x+1)-(x-\sqrt{\pi})(1)}{x^2+2x+1}$
and then simplified and got to $\frac{1+\sqrt{\pi}}{x^2+2x+1}$
and then I think i equal this to $\frac{1}{1+\sqrt{\pi}}$ because it's perpendicular... and then I don't know what to do. Do they cross out to be $\frac{1}{x^2+2x+1}$?Thank you!
 
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Cross multiplying would be the next step...
 

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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