18. [-/2 Points] DETAILS MY NOTES SCALCET9 3.4.061. A graphing calculator is recommended. Let $$ y=\frac{10}{1+\mathrm{e}^{-x}} $$. (a) Find an equation of the tangent line to the curve $$ y $$ at the point ( 0,5 ). (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.

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**Step 1. Calculate the derivative $$ y'(x) $$ of the function**

The function is given by  
$$
y = \frac{10}{1 + e^{-x}}.
$$  
We can rewrite it as  
$$
y = 10(1 + e^{-x})^{-1}.
$$

Differentiate using the chain rule. Let  
$$
u(x) = 1 + e^{-x},
$$  
so that  
$$
y = 10u^{-1}.
$$

Differentiate with respect to $$ x $$:
$$
y'(x) = 10 \cdot (-1) \cdot u^{-2} \cdot u'(x) = -10 \cdot u^{-2} \cdot u'(x).
$$

Now, differentiate $$ u(x) $$:
$$
u'(x) = \frac{d}{dx}(1 + e^{-x}) = -e^{-x}.
$$

Substitute $$ u'(x) $$ into the derivative:
$$
y'(x) = -10 \cdot \frac{1}{(1 + e^{-x})^2} \cdot (-e^{-x}) = \frac{10e^{-x}}{(1+e^{-x})^2}.
$$

**Step 2. Evaluate $$ y'(x) $$ at $$ x=0 $$**

Plug $$ x = 0 $$ into the derivative:
$$
y'(0) = \frac{10e^0}{(1+e^0)^2} = \frac{10 \cdot 1}{(1+1)^2} = \frac{10}{4} = \frac{5}{2}.
$$

So, the slope of the tangent line at $$ x = 0 $$ is  
$$
m = \frac{5}{2}.
$$

**Step 3. Write the equation of the tangent line**

The point on the curve at $$ x=0 $$ is given as $$ (0,5) $$.  
Using the point-slope form:
$$
y - y_1 = m(x - x_1),
$$
with $$ (x_1, y_1) = (0, 5) $$ and $$ m = \frac{5}{2} $$, we get:
$$
y - 5 = \frac{5}{2}(x - 0).
$$

Thus, the equation of the tangent line simplifies to:
$$
y = \frac{5}{2}x + 5.
$$

**Step 4. Graphing the curve and the tangent line**

To illustrate the result, graph the function
$$
y = \frac{10}{1+e^{-x}}
$$
and the tangent line
$$
y = \frac{5}{2}x + 5
$$
on the same coordinate plane. The graphing calculator or software (e.g., Desmos, GeoGebra) should display a sigmoid curve for the function and a straight line tangent to the curve at the point $$ (0,5) $$.
**Part (a): Equation of the Tangent Line** At the point $$ (0,5) $$, the tangent line to the curve $$ y = \frac{10}{1 + e^{-x}} $$ is $$ y = \frac{5}{2}x + 5. $$ **Part (b): Graphing the Curve and Tangent Line** Graph both the curve $$ y = \frac{10}{1 + e^{-x}} $$ and the tangent line $$ y = \frac{5}{2}x + 5 $$ on the same coordinate plane to visualize their intersection at $$ (0,5) $$.

[-/2 Points] DETAILS MY NOTES SCALCET9 3.4.061.
A graphing calculator is recommended.
Let .
(a) Find an equation of the tangent line to the curve at the point ( 0,5 ).
(b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.

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[-/2 Points] DETAILS MY NOTES SCALCET9 3.4.061. A graphing calculator is recommended. Let . (a) Find an equation of the tangent line to the curve at the point ( 0,5 ). (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.