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 $y = x - 1$.

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.