Answer:
[tex]\textsf{a)} \quad g'(x)=2x-4[/tex]
[tex]\textsf{b)} \quad y=2x[/tex]
[tex]\textsf{c)} \quad x=5[/tex]
Step-by-step explanation:
Given function:
[tex]g(x) = x^2-4x+9[/tex]
Differentiate the given function with respect to x:
[tex]\begin{aligned}g(x) & = x^2-4x+9\\\implies g'(x) & = 2 \cdot x^{2-1}-1 \cdot 4x^{1-1}+0\\& = 2x^{1}-4x^{0}\\& = 2x-4(1)\\& = 2x-4 \end{aligned}[/tex]
The y-value of the function at x = 3:
[tex]\begin{aligned}\implies g(3) & = (3)^2-4(3)+9\\& = 9-12+9\\& = -3+9\\& = 6\end{aligned}[/tex]
Therefore: (3, 6).
To find the slope of the function at x = 3, substitute x = 3 into the differentiated function:
[tex]\begin{aligned}\implies g'(x) & = 2(3)-4\\& = 6-4\\& = 2\end{aligned}[/tex]
Therefore, the slope of the function at x = 3 is 2.
Substitute the found slope and point (3, 6) into the point-slope form of linear equation to find the equation of the tangent line of g(x) at x = 3:
[tex]\begin{aligned} y-y_1 & =m(x-x_1)\\\implies y-6 & = 2(x-3)\\y-6 & = 2-6\\y & = 2x-6+6\\y&=2x\end{aligned}[/tex]
To find the value of x for which the slope of the graph is equal to 6, set the differentiated function to 6 and solve for x:
[tex]\begin{aligned}g'(x) & = 6\\\implies 2x-4 & = 6\\2x & = 6+4\\2x & = 10\\x & = \dfrac{10}{2}\\x & = 5\end{aligned}[/tex]