MHB Find the equation of the tangent line pt 3 (Need someone to check my work)

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The discussion focuses on finding the equation of the tangent line for the function y = e^(3x + cos x) at x = 0. The calculated point on the curve at x = 0 is y1 = e. The derivative of the function is y' = e^(3x + cos x) * (3 - sin x), leading to a slope (m) of 3e at that point. The equation of the tangent line is derived as y - e = 3e(x - 0). The participants confirm the calculations are correct, alleviating concerns about the value of e.
shamieh
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find the equation of the tangent line.$$y = e^{3x + cos x}$$ @ x = 0

$$
y1 = e^{3(0) + cos(0)} = e^1 = e$$
$$y1 = e$$

$$y = e^{3x+cos x}$$
$$y' = e^{3x + cos x} * (3 - sinx)$$

$$m = 3e$$

$$
y - e = 3e(x - 0) $$
 
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shamieh said:
find the equation of the tangent line.$$y = e^{3x + cos x}$$ @ x = 0

$$
y1 = e^{3(0) + cos(0)} = e^1 = e$$
$$y1 = e$$

$$y = e^{3x+cos x}$$
$$y' = e^{3x + cos x} * (3 - sinx)$$

$$m = 3e$$

$$
y - e = 3e(x - 0) $$
Looks good to me!

-Dan
 
Thank God! Was so worried about the $$e^1 = e $$
 
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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