Finding the Tangent line of g(x) = (16/x) - 4(sq rt of x) at x = 4

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mikagurl

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Hi all. I have to find the equation of the tangent line of g(x) = (16/x) - 4(square root of x) at x = 4.

I started by plugging in the 4 for the original function which gave me (16/4) - 4(square root of 4). I ended up with 4 - 8 which is g = -2. Then I had to find the derivative of g(x) = (16/x) - 4(square root of x).

I moved the x of the first term to the numerator which is 16x^-1. Then I changed the square root to a 1/2 power. So now for the complete derivative I have g'(x) = (-16x^-2) - 2(x)^(-1/2). I changed the derivative to (-16/x^2) - (2/square root x) so I could work with positive exponents only. The derivative at x = 4 is -2.

The equation of the tangent line is f(a) + f'(a)(x - a). So my line should be -4 - 2(x - 4). However, the answer is y = -2x + 4. I see that they distributed the -2. But what happened to the 4 that I have highlighted? Thanks!
 
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Yes and thanks. My mistake. So the tangent line should still be y=-4-2(x-4). For some reason the answer is y=-2x+4. What happened to the -4(the first term)? I see that the -2 was distributed. But I do not see how the -4 just went away.
 
mikagurl said:
The equation of the tangent line is f(a) + f'(a)(x - a). So my line should be -4 - 2(x - 4). However, the answer is y = -2x + 4. I see that they distributed the -2. But what happened to the 4 that I have highlighted? Thanks!
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What do you get when you distribute the -2 through the parentheses? What do you get when you combine the -4 and the +8? :wink:
 
This is a "Calculus" problem but you have two errors in integer arithmetic. That should concern you.

I doubt that it is because you do not know arithmetic! So I suspect you are simply not paying attention to what you are doing.
 
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It took me years to start "paying attention" to the problems sufficiently. Well, actually, I am still working on it!
 
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