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can anyone tell me the Use the linear approximation of the function f(x)=arctan(e3x) at x=0 to estimate the value of f(0.01).

Calculus1
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Find the equation of the tangent line of f(x) at x = 0 and evaluate f(0.01) using the tangent line. Do you know how to perform these steps? @fozia
im trying
btw, is that e^3x?

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yes it is
Ok let's first find the equation of the tangent line. To do this, I will need the slope of the line and a point. To get a point, we find f(0):\[\bf f(0)=\arctan(1)=\frac{ \pi }{ 4}\]So the tangent goes through the point \(\bf (0, \frac{\pi}{4})\). Now to find the slope of the tangent, we evaluate f'(x) at x = 0:\[\bf f'(x)=\frac{ 3e^{3x} }{ 1+e^{6x} } \rightarrow f'(0)=\frac{ 3e^0 }{ 1+e^0 }=3\]So now we have a point and the slope of tangent line. We will use the slope-intercept form (you can use point-slope form as well but I find slope-intercept form easier) to get the tangent line's equation:\[\bf y = mx+b \rightarrow y = 3x + b\]Plug in the point for 'x' and 'y':\[\bf \frac{\pi}{4}=3(0)+b \implies b = \frac{\pi}{4}\]So the equation of the tangent line is:\[\bf g(x)=3x+\frac{\pi}{4}\]Here I called the tangent line g(x). Now to find the linear approximation of f(0.01), we plug in x = 0.01 in to our equation of the tangent line and evaluate:\[\bf g(0.01)=3(0.01) + \frac{\pi}{4} \approx 0.815\]Therefore: \(\bf f(0.01) \approx 0.815\) @fozia
oh grt thank you so much

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