Question - Finding Tangent Line Equation Using Calculus
Solution:
\begin{align*}g(x) &= \frac{16}{x} - 4\sqrt{x}\\g'(x) &= -\frac{16}{x^2} - \frac{4}{2\sqrt{x}}\\g'(x) &= -\frac{16}{x^2} - \frac{2}{\sqrt{x}}\\g'(-4) &= -\frac{16}{(-4)^2} - \frac{2}{\sqrt{-4}}\\g'(-4) &= -\frac{16}{16} - \frac{2}{2i}\\g'(-4) &= -1 - \frac{1}{i}\\g'(-4) &= -1 + i\\g(-4) &= \frac{16}{-4} - 4\sqrt{-4}\\g(-4) &= -4 - 4 \cdot 2i\\g(-4) &= -4 - 8i\\\text{Point-slope form: } y - y_1 &= m(x - x_1)\\y - (-4 - 8i) &= (-1 + i)(x - (-4))\\y + 4 + 8i &= (-1 + i)(x + 4)\\y &= (-1 + i)x - 4 + 1 - 4i - 8i\\y &= (-1 + i)x - 3 - 12i\\\text{Tangent line equation: } y &= (-1 + i)x - 3 - 12i\end{align*}
Tutor Solution:
To find the tangent line to the function $$ g(x) = \frac{16}{x} - 4\sqrt{x} $$ at $$ x = 4 $$, we need to determine the slope of the tangent line at this point and use the point-slope form of a line to write the equation. Here are the steps to do so:1. Find the derivative $$ g'(x) $$ to determine the slope of the tangent line. Differentiate $$ g(x) $$ using the quotient rule and the power rule:\[ g'(x) = \frac{d}{dx}\left(\frac{16}{x}\right) - \frac{d}{dx}(4\sqrt{x}) \]\[ g'(x) = \frac{d}{dx}\left(16x^{-1}\right) - \frac{d}{dx}\left(4x^{1/2}\right) \]\[ g'(x) = -16x^{-2} - 4 \cdot \frac{1}{2}x^{-1/2} \]\[ g'(x) = -\frac{16}{x^2} - \frac{2}{\sqrt{x}} \]2. Evaluate the derivative at $$ x = 4 $$ to find the slope of the tangent line at that point:\[ g'(4) = -\frac{16}{(4)^2} - \frac{2}{\sqrt{4}} \]\[ g'(4) = -\frac{16}{16} - \frac{2}{2} \]\[ g'(4) = -1 - 1 \]\[ g'(4) = -2 \]So, the slope of the tangent line at $$ x = 4 $$ is -2.3. Find the y-coordinate of the point where $$ x = 4 $$ by evaluating $$ g(4) $$: \[ g(4) = \frac{16}{4} - 4\sqrt{4} \]\[ g(4) = 4 - 4(2) \]\[ g(4) = 4 - 8 \]\[ g(4) = -4 \]So the point is $$ (4, -4) $$.4. Use the point-slope form of the equation of a line to write the equation of the tangent line:Let $$ (x_1,y_1) $$ be $$ (4, -4) $$ and $$ m $$ be the slope which is -2. The point-slope form is:\[ y - y_1 = m(x - x_1) \]\[ y - (-4) = -2(x - 4) \]\[ y + 4 = -2x + 8 \]\[ y = -2x + 8 - 4 \]\[ y = -2x + 4 \]Hence, the equation of the tangent line to $$ g(x) $$ at $$ x = 4 $$ is $$ y = -2x + 4 $$.
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