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.
karush
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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. :)
 

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