Question

Pls help w this pretty urgent

Answer 1

`\textsf{a)} \quad g'(x)=2x-4`

`\textsf{b)} \quad y=2x`

`\textsf{c)} \quad x=5`

Explanation:

Given function:

`g(x) = x^2-4x+9`

Part (a)

Differentiate the given function with respect to x:

`\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}`

Part (b)

The y-value of the function at x = 3:

`\begin{aligned}\implies g(3) & = (3)^2-4(3)+9\\& = 9-12+9\\& = -3+9\\& = 6\end{aligned}`

Therefore: (3, 6).

To find the slope of the function at x = 3, substitute x = 3 into the differentiated function:

`\begin{aligned}\implies g'(x) & = 2(3)-4\\& = 6-4\\& = 2\end{aligned}`

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:

`\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}`

Part (c)

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:

`\begin{aligned}g'(x) & = 6\\\implies 2x-4 & = 6\\2x & = 6+4\\2x & = 10\\x & = (10)/(2)\\x & = 5\end{aligned}`