Question

Let f(x) = 12x^2 -3x (a) Use the limit process to find the slope of the line tangent to the graph of f at x = 3.Slope at x = 3:

Answer 1

Solution

$\begin{gathered} f(x)=12x^2-3x \\ f(x+\Delta x)=12(x+\Delta x)^2-3(x+\Delta x) \\ \\ \text{Subtract }f(x)\text{ from the second equation} \\ \\ f(x+\Delta x)-f(x)=12(x+\Delta x)^2-3(x+\Delta x)-(12x^2-3x) \\ \text{Expand the right-hand side and simplify} \\ \\ f(x+\Delta x)-f(x)=12(x^2+2x\Delta x+\Delta x^2)-3x-3\Delta x-12x^2+3x \\ \\ f(x+\Delta x)-f(x)=12x^2+24x\Delta x+12\Delta x^2-3x-3\Delta x-12x^2+3x \\ \\ \text{Collect like terms} \\ \\ f(x+\Delta x)-f(x)=12x^2-12x^2+24x\Delta x-3\Delta x+3x-3x+12\Delta x^2 \\ \\ f(x+\Delta x)-f(x)=24x\Delta x-3\Delta x+12\Delta x^2 \\ \\ \text{Divide both sides by }\Delta x \\ \frac{f(x+\Delta x)-f(x)}{\Delta x}=\frac{24x\Delta x-3\Delta x+12\Delta x^2}{\Delta x}=\frac{24x\Delta x}{\Delta x}-\frac{3\Delta x}{\Delta x}+\frac{12\Delta x^2}{\Delta x} \\ \\ \Delta x\text{ crosses out} \\ \\ \frac{f(x+\Delta x)-f(x)}{\Delta x}=24x-3+12\Delta x \\ \\ as\text{ }\Delta x\to0 \\ \\ \frac{f(x+\Delta x)-f(x)}{\Delta x}=24x-3 \end{gathered}$

- Now that we have the expression for the limit, let us find the slope at x = 3

- The slope is given by

$\begin{gathered} \text{slope}=24x-3 \\ x=3 \\ \text{slope}=24(3)-3 \\ \text{slope}=69 \end{gathered}$

Final Answer

The slope at x = 3, is 69