Consider the function $$ f(x)=x^{3}+3 $$ and the point $$ (1,4) $$. (a) Find the slope of the tangent to the graph of $$ f(x) $$ at any point. (b) Find the slope of the tangent at the given point. (c) Write the equation of the line tangent to the graph of $$ f(x) $$ at the given point. (Let $$ x $$ be the independent variable and $$ y $$ be the dependent variable.) (d) Graph both $$ f(x) $$ and its tangent line.

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Let's analyze the function $$ f(x) = x^{3} + 3 $$ and the point $$ (1, 4) $$ step by step.

### (a) **Finding the Slope of the Tangent at Any Point**

To find the slope of the tangent to the graph of $$ f(x) $$ at any point, we need to compute the derivative of $$ f(x) $$ with respect to $$ x $$.

$$ f(x) = x^{3} + 3 $$

Taking the derivative:

$$
f'(x) = \frac{d}{dx}(x^{3}) + \frac{d}{dx}(3) = 3x^{2} + 0 = 3x^{2}
$$

**Answer:** The slope of the tangent at any point $$ x $$ is $$ f'(x) = 3x^{2} $$.

### (b) **Finding the Slope of the Tangent at the Given Point $$ (1, 4) $$**

Using the derivative from part (a), we can find the slope at $$ x = 1 $$:

$$
f'(1) = 3(1)^{2} = 3 \times 1 = 3
$$

**Answer:** The slope of the tangent at the point $$ (1, 4) $$ is $$ 3 $$.

### (c) **Writing the Equation of the Tangent Line at $$ (1, 4) $$**

To write the equation of the tangent line, we'll use the point-slope form of a line:

$$
y - y_{1} = m(x - x_{1})
$$

Where:
- $$ m $$ is the slope of the tangent.
- $$ (x_{1}, y_{1}) $$ is the point of tangency.

From part (b):
- $$ m = 3 $$
- $$ (x_{1}, y_{1}) = (1, 4) $$

Plugging these values in:

$$
y - 4 = 3(x - 1)
$$

Simplifying to slope-intercept form ($$ y = mx + b $$):

$$
y - 4 = 3x - 3 \
y = 3x - 3 + 4 \
y = 3x + 1
$$

**Answer:** The equation of the tangent line at $$ (1, 4) $$ is $$ y = 3x + 1 $$.

### (d) **Graphing $$ f(x) $$ and Its Tangent Line**

While I can't provide a visual graph here, I can describe how you can graph both the function and its tangent line:

1. **Graphing $$ f(x) = x^{3} + 3 $$:**
   - This is a cubic function shifted upward by 3 units.
   - It has a point of inflection at $$ x = 0 $$.
   - Plot points for various $$ x $$ values to see the shape, such as:
     - $$ f(-2) = (-2)^3 + 3 = -5 $$
     - $$ f(-1) = (-1)^3 + 3 = 2 $$
     - $$ f(0) = 0 + 3 = 3 $$
     - $$ f(1) = 1 + 3 = 4 $$
     - $$ f(2) = 8 + 3 = 11 $$

2. **Graphing the Tangent Line $$ y = 3x + 1 $$:**
   - This is a straight line with a slope of 3 and a y-intercept at $$ (0, 1) $$.
   - Plot two points to draw the line:
     - When $$ x = 0 $$, $$ y = 1 $$.
     - When $$ x = 1 $$, $$ y = 4 $$ (which coincides with the point of tangency).

3. **Combining Both Graphs:**
   - Plot the cubic curve and the straight tangent line on the same set of axes.
   - The tangent line $$ y = 3x + 1 $$ will touch the curve $$ f(x) = x^{3} + 3 $$ exactly at the point $$ (1, 4) $$ and have the same slope there.

**Visual Representation:**

If you plot both on graphing software or graph paper:

- The cubic curve will pass through the point $$ (1, 4) $$.
- The straight line $$ y = 3x + 1 $$ will just touch the cubic at $$ (1, 4) $$ without crossing it, illustrating that it's indeed a tangent at that point.

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**Summary of Answers:**

(a) The slope of the tangent at any point $$ x $$ is $$ f'(x) = 3x^{2} $$.

(b) The slope of the tangent at $$ (1, 4) $$ is $$ 3 $$.

(c) The equation of the tangent line at $$ (1, 4) $$ is $$ y = 3x + 1 $$.

(d) [Graph Description as above.]
(a) The slope of the tangent at any point $$ x $$ is $$ 3x^{2} $$. (b) The slope of the tangent at $$ (1, 4) $$ is $$ 3 $$. (c) The equation of the tangent line at $$ (1, 4) $$ is $$ y = 3x + 1 $$. (d) The graph shows the cubic function $$ f(x) = x^{3} + 3 $$ and its tangent line $$ y = 3x + 1 $$ at the point $$ (1, 4) $$.

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