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Find an equation of the tangent line to the graph of the

function f at the specified point. f(x) = 2 cos x - 3 sin x; (π,
-2) (Give your answer in the slope-intercept form.)
Find an equation of the tangent line to the graph of the function \( f \) at the specified point. \[ f(x)=2 \cos x-3 \sin x ; \quad(\pi,-2) \] (Give your answer in the slope-intercept form.)

User Fgb
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Answer:

To find the equation of the tangent line to the graph of the function \(f(x) = 2\cos(x) - 3\sin(x)\) at the point \((\pi, -2)\), you'll need to find both the slope of the tangent line and the y-intercept.

The slope of the tangent line at a point can be found by taking the derivative of the function and then evaluating it at the given x-coordinate.

First, let's find the derivative of \(f(x)\):

\[f'(x) = \frac{d}{dx}(2\cos(x) - 3\sin(x))\]

Using the derivative rules:

\[f'(x) = -2\sin(x) - 3\cos(x)\]

Now, we need to find the slope of the tangent line at \(x = \pi\):

\[f'(\pi) = -2\sin(\pi) - 3\cos(\pi) = 0 - (-3) = 3\]

So, the slope of the tangent line at \((\pi, -2)\) is 3.

Now that we have the slope (\(m = 3\)), we can use the point-slope form of a linear equation to find the equation of the tangent line:

\[y - y_1 = m(x - x_1)\]

Where \((x_1, y_1)\) is the given point on the curve \((\pi, -2)\). Plugging in the values:

\[y - (-2) = 3(x - \pi)\]

Now, simplify the equation:

\[y + 2 = 3x - 3\pi\]

To express it in slope-intercept form (\(y = mx + b\)), isolate \(y\):

\[y = 3x - 3\pi - 2\]

So, the equation of the tangent line to the graph of \(f(x) = 2\cos(x) - 3\sin(x)\) at the point \((\pi, -2)\) is:

\[y = 3x - 3\pi - 2\]

Explanation:

User Michele Da Ros
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