Question

I need help with a calculus problem. My instructor wants me to sketch the graph of a function that is differentiable and has specific derivative values at x=-3 and x=1. Can you find an example of this type of problem and show me, step by step, how to sketch the graph?

Answer 1

Certainly! Let's start with a function that is differentiable. In our case, we'll define a simple polynomial function, but there are many forms you could use (for instance, trigonometric, exponential, etc.).

### Step 1: Define the function and its derivative

Let's say we have a differentiable function \( f(x) \). We want to specify values for its derivative at two points:

- \( f'(-3) = 2 \) (the slope of the tangent line at \( x = -3 \))
- \( f'(1) = -1 \) (the slope of the tangent line at \( x = 1 \))

### Step 2: Choose an appropriate function

A polynomial function will work well here. Let's consider a cubic function of the form:

\[
f(x) = ax^3 + bx^2 + cx + d
\]

### Step 3: Find the derivative

The derivative of \( f \) is:

\[
f'(x) = 3ax^2 + 2bx + c
\]

### Step 4: Set up equations for the derivative

Using the derivative, we can set up the equations based on the specific points we are interested in:

1. For \( x = -3 \):
\[
f'(-3) = 3a(-3)^2 + 2b(-3) + c = 2
\]
Simplifying,
\[
27a - 6b + c = 2 \quad \text{(Equation 1)}
\]

2. For \( x = 1 \):
\[
f'(1) = 3a(1)^2 + 2b(1) + c = -1
\]
Simplifying,
\[
3a + 2b + c = -1 \quad \text{(Equation 2)}
\]

### Step 5: Choose values for coefficients

Since we have two equations and three unknowns (a, b, c), we can choose a value for one of the coefficients. Let's set \( a = 1 \) (you could choose different values, but this is a common choice):

- Plugging \( a = 1 \) into the equations gives:

From Equation 1:
\[
27(1) - 6b + c = 2 \implies 27 - 6b + c = 2 \implies c = 6b - 25
\]

From Equation 2:
\[
3(1) + 2b + c = -1 \implies 3 + 2b + c = -1
\]
Substituting \( c \) from the first equation:
\[
3 + 2b + (6b - 25) = -1 \implies 8b - 22 = -1 \implies 8b = 21 \implies b = \frac{21}{8}
\]

Now, find \( c \):
\[
c = 6\left(\frac{21}{8}\right) - 25 = \frac{126}{8} - 25 = \frac{126 - 200}{8} = \frac{-74}{8} = -\frac{37}{4}
\]

### Step 6: Write the final function

So, we have:
- \( a = 1 \)
- \( b = \frac{21}{8} \)
- \( c = -\frac{37}{4} \)

Thus, our function is:
\[
f(x) = x^3 + \frac{21}{8}x^2 - \frac{37}{4}x + d
\]

We can set \( d \) to any value (let's choose \( d = 0 \) for simplicity):
\[
f(x) = x^3 + \frac{21}{8}x^2 - \frac{37}{4}x
\]

### Step 7: Sketch the graph

1. **Identify critical points**: Use the information on the derivative to note that the function has a positive slope at \( x = -3 \) and a negative slope at \( x = 1 \).
2. **Calculate values at specific points**: You may want to evaluate \( f(-3) \) and \( f(1) \) to see the height of the function at these critical points.
3. **Check concavity**: Since it’s a cubic function, it will change concavity. You might also find the second derivative to analyze this further.
4. **Plot points**: Sketch the graph using calculated points and noting the sign of the derivative at specified intervals around critical points.

This should give you a solid function graphically consistent with your derivative conditions!