Curve Slope #1

$f(x) = 3x^{2} + 6x + 3$

$f'(x) = 6x + 6$

Since the derivative of a function gives us the slope of the line tangent to the function at a certain point on the graph, substituting $x=3$ we get:

$f'(x) = 18 + 6 = \boxed{24}$

The first thing that should come to mind when thinking of finding the slope of a curve at a certain point (finding the slope of the line tangent to a point on a curve) is derivatives. Note that by deriving $f'(x)$ , we can plug in $x=3$ to find the slope of the line tangent to $f(x)=3x^2+6x+3$ at $x=3$ , as derivatives are a measure of the rate of instantaneous change, or the sensitivity of a function to change of its input.

Using the sum rule (if $f(x)=g(x)+h(x)+...$ , then $f'(x)=g'(x)+h'(x)+...$ ) and power rule (if $f(x)=ax^b$ , then $f'(x)=abx^{b\,-\,1}$ ), we can find the derivative of $f(x)$ without needing to solve $\displaystyle\lim_{\Delta x\to 0}{{f(x+\Delta x) - f(x)}\over{\Delta x}}$ for $f(x)=3x^2+6x+3$ .

Given:

• $f(x)=j(x)+k(x)+l(x)$

• $j(x)=3x^2$

• $k(x)=6x$

• $l(x)=3$

Then $f'(x)=j'(x)+k'(x)+l'(x)$ .

• $j'(x)=3\times 2x^{2\,-\,1}=6x$

• $k'(x)=6$

• $l'(x)=0$

• $f'(x)=j'(x)+k'(x)+l'(x)=6x+6$

We can then plug in $x=3$ , which gives: $6(3)+6=\boxed{24}$ , which must be the slope of the line tangent to $f(x)=3x^2+6x+3$ at $x=3$ .