Let f(x) be a continuous twice differentiable functoon with f(4) =17, f’(4)=-3, f’’(4)=2. Write an equation of the line tangent f at the point where x=4

To write the equation of the tangent line to the function \( f(x) \) at the point where \( x = 4 \), we can use the point-slope form of a line equation:

\[
y - y_1 = m(x - x_1)
\]

Where:
- \( (x_1, y_1) \) is the point on the function.
- \( m \) is the slope of the tangent line, which is given by \( f'(x_1) \).

We know the following information from the problem:
- \( f(4) = 17 \) gives us the point \( (4, 17) \).
- \( f'(4) = -3 \) gives us the slope of the tangent line at \( x = 4 \).

Using the point-slope form:

\[
y - 17 = -3(x - 4)
\]

This simplifies to:

\[
y = -3(x - 4) + 17
\]

\[
y = -3x + 12 + 17
\]

\[
y = -3x + 29
\]

Thus, the equation of the tangent line is:

\[
y = -3x + 29
\]

Would you like more details or have any questions about this?

Here are 5 related questions for further practice:
1. How would the tangent line change if \( f'(4) = 5 \) instead of \( -3 \)?
2. What is the significance of \( f''(4) = 2 \) in this context?
3. How can we use the second derivative to analyze the concavity at \( x = 4 \)?
4. If \( f(x) \) had a higher-order derivative at \( x = 4 \), how could we approximate \( f(x) \)?
5. Can you find the equation of the normal line to \( f(x) \) at \( x = 4 \)?

**Tip**: The second derivative \( f''(x) \) provides information about the concavity of the function, which helps in understanding the behavior of the graph around the point of tangency.

Let f(x) be a continuous twice differentiable function with f(4) =17, f’(4)=-3, f’’(4)=2. Write an equation of the line tangent to f at the point where x=4.

Learn how to find the equation of the tangent line to the function f(x) at x=4 using its first derivative. This calculus problem involves concepts like derivatives and point-slope form, perfect for students in advanced calculus courses.

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