MHB 10) AP Calculus linear functions

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The discussion focuses on finding the equations of tangent lines for the functions f and g at x=0. For function f, the tangent line is derived using the point-slope form, resulting in the equation y - 2 = -3(x - 0). For function g, the derivative g'(x) is calculated at x=0, yielding a slope of 6, leading to the tangent line equation y - 4 = 6(x - 0). Participants express uncertainty about the calculations related to g, specifically regarding the slope of its tangent line. The conversation emphasizes the importance of correctly determining derivatives to find tangent lines.
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$\textbf{10)} \\
f(x)\text{ is continuous at all } \textit{x}
\\
\displaystyle
f(0)=2, \, f'(0)=-3,\, f''(0)=0 $
$\text{let} \textbf{ g }
\text{be a function whose derivative is given by}\\
\displaystyle g'(x)=e^{-2 x} (3f(x))+2f'(x)
\text{ for all x}\\$
$\text{a) write an equation of the line tangent to the graph of f at the point where } $ $x=0$
$\displaystyle y-2 = -3(x-0) \\$
$\text{b) given that } \displaystyle g(0)=4, \\
\text{ write an equation of the line tangent of}
$g$
\text{at the point where }
\textit{x=0} \\$
$\displaystyle m_g=g'(0)=(1)[(3\cdot2)+2(0)]=6 \\
y-4=6(x-0)$

Was not sure about $g$ ??
 
Last edited:
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For part b), check the slope of the tangent line. :)
 
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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