Question

(a) find the slope m of the tangent to the curve y = \\(\frac{5}{\sqrt{x}}\\) at the point where x = a > 0.

m =

(b) find equations of the tangent line at the point (1, 5).

y =

find equations of the tangent line at the point \\((4, \frac{5}{2})\\).

y =

(c) graph the curve and both tangents on a common screen.

Explanation:

Step1: Rewrite the function

Rewrite $y = \frac{5}{\sqrt{x}}$ as $y = 5x^{-\frac{1}{2}}$.

Step2: Find the derivative

Using the power - rule $(x^n)'=nx^{n - 1}$, we have $y'=5\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-\frac{5}{2}x^{-\frac{3}{2}}=-\frac{5}{2x^{\frac{3}{2}}}$.

Step3: Find the slope at $x = a$

Substitute $x = a$ into the derivative. The slope $m$ of the tangent at $x = a>0$ is $m=-\frac{5}{2a^{\frac{3}{2}}}$.

Step4: Find the slope at $(1,5)$

Substitute $x = 1$ into $y'$. When $x = 1$, $y'=-\frac{5}{2\times1^{\frac{3}{2}}}=-\frac{5}{2}$. Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(1,5)$ and $m = -\frac{5}{2}$, we get $y-5=-\frac{5}{2}(x - 1)$. Expand it: $y-5=-\frac{5}{2}x+\frac{5}{2}$, so $y=-\frac{5}{2}x+\frac{5}{2}+5=-\frac{5}{2}x+\frac{15}{2}$.

Step5: Find the slope at $(4,\frac{5}{2})$

Substitute $x = 4$ into $y'$. When $x = 4$, $y'=-\frac{5}{2\times4^{\frac{3}{2}}}=-\frac{5}{2\times8}=-\frac{5}{16}$. Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(4,\frac{5}{2})$ and $m = -\frac{5}{16}$, we have $y-\frac{5}{2}=-\frac{5}{16}(x - 4)$. Expand it: $y-\frac{5}{2}=-\frac{5}{16}x+\frac{5}{4}$, so $y=-\frac{5}{16}x+\frac{5}{4}+\frac{5}{2}=-\frac{5}{16}x+\frac{15}{4}$.

Answer:

(a) $m = -\frac{5}{2a^{\frac{3}{2}}}$
(b) $y=-\frac{5}{2}x+\frac{15}{2}$; $y=-\frac{5}{16}x+\frac{15}{4}$
(c) Graphing the curve $y = \frac{5}{\sqrt{x}}$ and the tangent lines $y=-\frac{5}{2}x+\frac{15}{2}$ and $y=-\frac{5}{16}x+\frac{15}{4}$ can be done using graphing software like Desmos or a graphing calculator. Enter the functions $y = \frac{5}{\sqrt{x}}$, $y=-\frac{5}{2}x+\frac{15}{2}$, and $y=-\frac{5}{16}x+\frac{15}{4}$ to visualize them.