Science:Math Exam Resources/Courses/MATH110/April 2017/Question 01 (b)/Solution 1

From the part (a), one way to define derivative is slope of tangent line. In order words, the slope of the tangent line of $g(x)$ at $x=a$ is $g'(a)$ .

This gives the tangent line formula of $g$ at a point $(a,g(a))$ as $y=g(a)+g'(a)(x-a)$.

For $g(x)=x^{2}+2x$ and $a=1$, we have

$g(1)=1^{2}+2\cdot 1=3$, and $g'(1)=4$ from $g'(x)=2x+2$.

Therefore, plugging these into the tangent line formula, the tangent line of $g$ at $x=1$ is

$y=g(1)+g'(1)(x-1)=3+4(x-1)=4x-1$

Answer: $\color {blue}y=4x-1$