Tangent lines Find an equation of the line tangent to the grap...

Question

Tangent lines Find an equation of the line tangent to the graph of f at the given point.

f(x) = sin⁻¹(x/4); (2,π/6)

Accepted Answer

Find an equation of the line tangent to the graph of F of X equals cosine inverse, X divided by 2, at the 0.1 pi divided by 3. We have 4 possible answers here, which are all linear equations based on the following 4. Y tangent line is equivalent to F at some point X not. Multiplied by x minus x not. Plus F of X not. Where our points X not, why not? Is going to be 1 pi divided by 3. Now to solve this, we need to find F and F both of X not. So let's first find FX not. Well, we know X is 1, emitting our equation is F1, multiplied by X minus 1. Plus F of 1 And F1 will be given by the derivative of cosine inverts. So, we will find. F of X, which for cosine inverse is 1. Divided by the square root of 1 minus X squared but negative. Now, this is actually a chain rule function based on the X divided by 2. So we will modify our derivative. We get -1 divided by the square root of 1 minus X divided by 2 squad. Which is then multiplied by out inside derivative. In this case, the derivative of X divided by 2 is just 1/2. No. We need to find F of X and not by plugging in our value. F of 1 is equivalent to -1 divided by the square root 1 minus 1/2 squared. Multiplied by 1/2. This gives us -1. Divided by the square root. Of 1 minus 1/4, multiplied by 1/2. Which gives us -1 divided by the square root, 3/4 multiplied by 1/2. Now, we can simplify a square root. To get -1 divided by. 2 Multiplied By the square root of 3 divided by 2. Moving the two from the 1/2 multiplier into the fraction. We get -1 divided by the square root of 3. Which by rationalizing our denominator gives us 23 divided by 3. So that is our F of 1. We now just need to find our F of one. Which we do by plugging in to our function. Cosine in verse. Of one half. No. We just see the value for cosine and reverse of 1/2. F of 1, then we could find by looking at a unit circle. But in general, We end up getting. The value pi divided by 3 in radiance. So now we can assemble all of this into our function. Y tangent is equivalent to F1. Multiplied by x minus 1 plus F of 1. We end up getting 2 of 3 divided by 3, multiplied by X minus 1. Plus pi divided by 3. We can now simplify by distributing to give us negative, square 3 divided by 3 X plus square 3 divided by 3, plus pi divided by 3. And we can finally simplify this, to get negative square to 3 divided by 3 X plus pi + +23. All divided by a 3. This is our equation and our answer, and we can see this corresponds. With answer B. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Tangent lines Find an equation of the line tangent to the graph of f at the given point.f(x) = sin−1(x/4); (2,π/6)

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Tangent lines Find an equation of the line tangent to the graph of f at the given point.
f(x) = sin⁻¹(x/4); (2,π/6)