Use calculus to show that the line y=8x + 10 is a tangent to the graph of the function f(x)=3x^3 - x + 4

To show that the line \( y = 8x + 10 \) is a tangent to the graph of the function \( f(x) = 3x^3 - x + 4 \), we need to satisfy two conditions:
1. The line and the function intersect at exactly one point (i.e., they have a common point).
2. The slope of the tangent to the function at that point is equal to the slope of the line.

### Step 1: Find the slope of the function \( f(x) \)
The slope of the tangent line to the graph of the function \( f(x) \) at any point is given by the derivative \( f'(x) \). Let's find the derivative:

\[
f(x) = 3x^3 - x + 4
\]
\[
f'(x) = \frac{d}{dx}(3x^3 - x + 4) = 9x^2 - 1
\]

### Step 2: Compare the slope of the tangent line to the slope of the given line
The slope of the line \( y = 8x + 10 \) is 8.

For the line to be tangent to the function, the slope of the tangent line to the function at some point \( x = c \) must equal 8. Therefore, we set the derivative equal to 8:

\[
f'(x) = 8
\]
\[
9x^2 - 1 = 8
\]
Solve for \( x \):

\[
9x^2 = 9
\]
\[
x^2 = 1
\]
\[
x = \pm 1
\]

So, the line could potentially be tangent at \( x = 1 \) or \( x = -1 \).

### Step 3: Find the points of intersection
Now, let's find the y-values of the function \( f(x) \) at \( x = 1 \) and \( x = -1 \):

1. For \( x = 1 \):
   \[
   f(1) = 3(1)^3 - 1 + 4 = 3 - 1 + 4 = 6
   \]
   The point on the function is \( (1, 6) \).

2. For \( x = -1 \):
   \[
   f(-1) = 3(-1)^3 - (-1) + 4 = -3 + 1 + 4 = 2
   \]
   The point on the function is \( (-1, 2) \).

Now, check if these points satisfy the equation of the line \( y = 8x + 10 \):

1. For \( x = 1 \):
   \[
   y = 8(1) + 10 = 18
   \]
   The point \( (1, 6) \) does not lie on the line \( y = 8x + 10 \).

2. For \( x = -1 \):
   \[
   y = 8(-1) + 10 = -8 + 10 = 2
   \]
   The point \( (-1, 2) \) does lie on the line \( y = 8x + 10 \).

### Conclusion
The line \( y = 8x + 10 \) is a tangent to the graph of the function \( f(x) = 3x^3 - x + 4 \) at the point \( (-1, 2) \), where the slope of the function equals the slope of the line, and they intersect at that point.

Would you like any further details or have any questions?

Here are some related questions to consider:
1. What is the general method to find the tangent line to any curve?
2. How do you determine if a function has more than one tangent line at a point?
3. How can you find the point of intersection between a curve and a line without solving explicitly?
4. What role does the derivative play in understanding the behavior of functions?
5. Can a line be tangent to a curve at multiple points?

**Tip:** When finding where a line is tangent to a curve, always check both the slope and intersection conditions.

Learn how to use calculus to demonstrate that the line y=8x + 10 is a tangent to the graph of the function f(x)=3x^3 - x + 4. Understand the conditions for tangency, including slope equality and intersection at a single point. Suitable for college-level calculus students.

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