Find an equation for the tangent to the curve at the given point: f(x)=4(sqrt(x))-x+3, (16,3)
do you know how to differentiate?
\[f(x)=4\sqrt{x}-x+3\]
Yes but not so well with radicals
remember that \(\sqrt{x}\) can be written as \(\displaystyle x^{\frac{1}{2}}\)
does that help?
so 2/sqrt(x)
that is the first term - yes
\[d/dx=2/\sqrt{x}-1\]
yup, although usually written out as:\[f'(x)=\frac{2}{\sqrt{x}}-1\]
so what do I do with coordinates (16,3)? Just substitute?
the differential will give you the value of the slope of the tangent at any point
so, since the differential involves terms only containing x, you would substitute x=16 to get the value of the slope of the tangent line
so 1/2 is the slope and 3 is the y intercept so the answer should be y=1/2x+3
that doesn't seem right to me
oohhhso 2/4x-16+3 the answer is 1/2x-11??
lets do this one step at a time to avoid mistakes. first what do you get for the value of the slope at (16,3) ?
2/sqrt(16)x =2/4x =1/2x
Or (1/2)x
no - remember the equation that tells us what the slope of the tangent line will be at any point is given by:\[f'(x)=\frac{2}{\sqrt{x}}-1\]so substitute x=16 into this equation and what do you get for the value of f'(x)?
-1/2
correct, so now you know the equation of the tangent line is:\[y=-\frac{1}{2}x +c\]
next, we know that this line must also pass through the point (16,3). So use this to calculate the value for c.
so I need to solve -(1/2)(16)+c=0?
or =3?
not =0, but =3
ok give me one sec to get this
remember it passes through the point (16,3) => x=16, y=3
that cant be right...-(1/2)(16)=-8 c=3-8 =-5 That is not an option. It has to be 3, -11, or 11
nevermind 11 :)
you made a mistake. remember the equation is:\[y=-\frac{1}{2}x +c\]
I think you know what I did :D
:)
Thank you so much for you help
yw :)
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