MHB Find the Tangent lines of the slopes of the three zeroes

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The discussion focuses on finding the tangent lines of the slopes at three identified zeroes of the function f(x) = 1 + (50sin(x))/(x^2 + 3) within the interval -5 < x < 5. The user has already determined the zeroes to be approximately 0.02, 3.16, and -3.12. The next step involves calculating the derivative of the function, with a suggestion to use the quotient rule for differentiation. The conversation emphasizes the importance of confirming the correct form of the equation before proceeding with the derivative. The user seeks assistance specifically with deriving the slopes of the tangent lines at the zeroes.
Nivetham
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1+50sinx/x^2+3
-5 < x < 5

3 zeroes: 0.02, 3.16, -3.12
Find the derivative and the slopes of the tangent lines.

I need help with the last part. I found out the three zeroes by adding and subtracting pi from the equation at top by setting it to zero.
Thank you!
 
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Hi Nivetham,

Welcome to MHB! :)

Is this your equation? $$f(x)=1+\frac{50 \sin(x)}{x^2}+3?$$ or is it $$f(x)=1+\frac{50 \sin(x)}{x^2+3}$$, or is it $$\frac{1+50 \sin(x)}{x^2+1}$$?

Jameson
 
Hi! It's the second equation. Thank you!
 
Nivetham said:
Hi! It's the second equation. Thank you!
Hello Nivetham,
Have you derivate the function?
Hint: Quotient rule

Regards,
 

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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