Struggling to understand this
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calculus problem. Think I know where to start, but am unsure of how
to finish it.$$y=F(x)=8\sqrt{x}-x+5,\\ (x,y)=(64,5)$$
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I take the derivative of the equation $8x^{1/2}-x^{-1} +5$ Then I plug in the $x$ value (I think)?
$8(64^{1/2})-64+5$ simplifies to $64-64+5=5$. slope$=5$ Correct?
3.Then I put it in the slope formula? and use the given points for the second $y$ and $x$ values?
$$y-y_1=m(x-x_1)\\ y-5=5(x-64)$$ simplifies to $y-5=5x-320$ which simplifies to $y=5x-315$. This is not an answer option and I am unsure of the steps I used or where I went wrong. Any help would be appreciated.
Best Answer
To find the equation of the tangent line to the function
$$F(x) = 8\sqrt{x} - x + 5$$
at the point $(64, 5)$, we find the derivative, evaluate it when $x = 64$, then find the equation of the line through the point $(64, 5)$ whose slope is equal to $F'(64)$.
\begin{align*} F(x) & = 8\sqrt{x} - x + 5\\ & = 8x^{\frac{1}{2}} - x + 5 \end{align*}
Differentiating yields
\begin{align*} F'(x) & = \frac{1}{2} \cdot 8x^{-\frac{1}{2}} - 1\\ & = \frac{4}{\sqrt{x}} - 1 \end{align*}
Evaluating the derivative when $x = 64$ yields
\begin{align*} F'(64) & = \frac{4}{\sqrt{64}} - 1\\ & = \frac{4}{8} - 1\\ & = \frac{1}{2} - 1\\ & = -\frac{1}{2} \end{align*}
Hence, the tangent line to the graph of $F(x)$ at the point $(64, 5)$ has slope $-1/2$. Using the point-slope equation $$y - y_0 = m(x - x_0)$$ with $(x_0, y_0) = (64, 5)$ and $m = -1/2$ yields
$$y - 5 = -\frac{1}{2}(x - 64)$$