Question

Let f(x) = 9x^2- 6x.(a) Use the limit process or derivative to find the slope of the line tangent to the graph of f at x = 3

Answer 1

Use the limit derivative to find slope:

$f(x)=9x^2-6x$

The limit definition of a derivative is given by:

$f(x)=\lim _{h\to0}\frac{f(x+h)-f(x)}{h}$

Here, the derivative is:

$\begin{gathered} f^1(x)=\lim _{h\to0}\frac{9(x+h)^2-6(x+h)-(9x^2-6x)}{h} \\ =\lim _{h\to0}\frac{9(x^2+2xh+h^2)-6x-6h-9x^2+6x}{h} \\ =\lim _{h\to0}\frac{9x^2+18xh+9h^2-6x-6h-9x^2+6x}{h} \\ =\lim _{h\to0}\frac{18xh+9h^2-6h}{h} \\ =\lim _{h\to0}\frac{18xh}{h}+\frac{9h^2}{h}-\frac{6h}{h} \\ =18x+\lim _{h\to0}\frac{9}{h}-6 \\ =18x-6 \end{gathered}$$f^1(x)=18x-6$

We have now found the gradient of the tangent at any point. All that's left is to evaluate it at x = 3

$\begin{gathered} f^1(x)=18x-6 \\ y=18(3)-6 \\ y=54-6 \\ y=48 \end{gathered}$

Hence the slope of the equation = 48