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I am stumped with this question.

Find the line tangent $f(t)=3\sin(2t)+5$ at the point where $t=\pi$.

You must first find the derivative of $f$ at $\pi$. Next, find the equation of the tangent line using the slope$=f′(π)$ and the point $P(\pi)$. Yep, you need to find $y$.

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  • $\begingroup$ Hello and welcome to Mathematics StackExchange! Do you know the equation for the tangent line of a function at some point? There is a general formula for that. $\endgroup$ – Markus Zetto May 5 '20 at 16:08
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The problem specifies the point $(t_1,y_1)=\left(\pi,f(\pi)\right)$ where $f(t)=3\sin(2t)+5$ and $y=f(t)$. The equation of a tangent line is given by

$$y-y_1=m(t-t_1).\tag{1}$$

You need to find $f(\pi)$ and $m=f'(\pi)$ and then substitute them into $(1)$.

Step 1: Find the derivative of $f$ at $\pi$.

$$f(t)=3\sin(2t)+5$$ $$f'(t)=6\cos(2t)$$ $$f'(\pi)=6\cos(2\pi)=6$$

Step 2: Find $f(\pi)$.

$$f(\pi)=3\sin(2\pi)+5=5$$

Therefore, the point is $(t_1,y_1)=(\pi,5)$ and the slope of the tangent line is $m=f'(\pi)=6$. Substitute these values into $(1)$ to find the equation of the tangent line.

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The derivative is the slope of the tangent line. If you plug in pi into the equation of the derivative you get the slope at the point pi: f’( pi ). If you have a point and a slope, you can use the point-slope formula to express a line. If you plug in pi in to the equation, you get the point ( pi, f( pi ) ).

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At $f'(π)=6cos(2π)=6$. So, the slope is positive.

So we get that the equation of the tangent line will be

$\frac{y-f(π)}{x-π} =6 \Rightarrow y=6x+(5-6π)$.

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