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Let's break down and solve each of the problems shown in the image.

### Problem 1:
> The gradient of the tangent line to the graph of \( f(x) = 2x^2 + kx - 3 \) at \( x = 1 \) is 1.5. Find the value of \( k \).

To solve this, follow these steps:
1. First, find the derivative of the function \( f(x) = 2x^2 + kx - 3 \). This will give us the slope of the tangent line at any point \( x \).
   \[
   f'(x) = \frac{d}{dx}(2x^2 + kx - 3) = 4x + k
   \]
2. At \( x = 1 \), the gradient (slope) is given as 1.5. Set \( f'(1) = 1.5 \):
   \[
   f'(1) = 4(1) + k = 1.5
   \]
   \[
   4 + k = 1.5
   \]
   \[
   k = 1.5 - 4 = -2.5
   \]
So, the value of \( k \) is \( -2.5 \).

### Problem 2:
> Write the equation of the tangent line to the graph of \( f(x) = x^3 \ln x \) at \( x = 1 \).

To solve this:
1. Find the derivative of the function \( f(x) = x^3 \ln x \) using the product rule. Let \( u(x) = x^3 \) and \( v(x) = \ln x \), so the derivative is:
   \[
   f'(x) = \frac{d}{dx}(x^3) \cdot \ln x + x^3 \cdot \frac{d}{dx}(\ln x)
   \]
   \[
   f'(x) = 3x^2 \ln x + x^3 \cdot \frac{1}{x}
   \]
   Simplifying:
   \[
   f'(x) = 3x^2 \ln x + x^2
   \]
2. Evaluate this derivative at \( x = 1 \):
   \[
   f'(1) = 3(1)^2 \ln(1) + (1)^2 = 0 + 1 = 1
   \]
   So, the slope of the tangent line at \( x = 1 \) is 1.
   
3. To find the equation of the tangent line, we also need the function value at \( x = 1 \):
   \[
   f(1) = (1)^3 \ln(1) = 1 \cdot 0 = 0
   \]
4. The equation of the tangent line is given by the point-slope form:
   \[
   y - f(1) = f'(1)(x - 1)
   \]
   Substituting \( f(1) = 0 \) and \( f'(1) = 1 \):
   \[
   y - 0 = 1(x - 1)
   \]
   \[
   y = x - 1
   \]
Thus, the equation of the tangent line is \( y = x - 1 \).

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Do you want further clarifications on any step or details?  
Here are 5 related questions to practice:

1. How would you find the tangent line equation if the function involved a different logarithmic or exponential term?
2. What is the meaning of the gradient (slope) in the context of a tangent line to a curve?
3. How does the value of \( k \) in Problem 1 affect the shape of the parabola?
4. What would the derivative of \( f(x) = x^n \ln x \) be for different powers of \( n \)?
5. How can you apply the chain rule when differentiating a composite function like \( (x^2 + 1) \ln x \)?

**Tip**: When dealing with tangent lines, always find both the derivative at the point and the function value at that same point for accuracy.

1. The gradient of the tangent line to the graph of f(x) = 2x^2 + kx - 3 at x = 1 is 1.5. Find the value of k.
2. Write the equation of the tangent line to the graph of f(x) = x^3 ln(x) at x = 1.

Learn how to find the value of k for the quadratic function f(x) = 2x^2 + kx - 3 by using the derivative and the gradient of the tangent line at x = 1. Additionally, find the equation of the tangent line to the graph of f(x) = x^3 ln(x) at x = 1 using the product rule. Suitable for students in Grades 11-12.

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