problem 2. suppose that y = 5 - 2x is the tangent line to y = f(x) at x = 3 and that y = 2 + 3x is the…

Explanation:

Step1: Find $f(3)$ and $f'(3)$

The tangent - line to $y = f(x)$ at $x = 3$ is $y=5 - 2x$. The slope of the tangent line is the derivative of the function at that point, so $f'(3)=-2$. Also, when $x = 3$, $y=f(3)=5-2\times3=-1$.

Step2: Find $g(3)$ and $g'(3)$

The tangent - line to $y = g(x)+x$ at $x = 3$ is $y = 2+3x$. The slope of the tangent line to $y = g(x)+x$ at $x = 3$ is $3$. Since the derivative of $y = g(x)+x$ is $y'=g'(x)+1$, then $g'(3)+1 = 3$, so $g'(3)=2$. When $x = 3$, $g(3)+3=2 + 3\times3$, so $g(3)=2 + 9-3=8$.

Step3: Find $h(3)$

We know that $h(x)=f(x)g(x)+x^{2}$. Then $h(3)=f(3)g(3)+3^{2}=(-1)\times8 + 9=1$.

Step4: Find $h'(x)$ using the product rule

The product rule states that $(uv)'=u'v+uv'$, where $u = f(x)$ and $v = g(x)$. So $h'(x)=f'(x)g(x)+f(x)g'(x)+2x$.

Step5: Find $h'(3)$

Substitute $x = 3$ into $h'(x)$: $h'(3)=f'(3)g(3)+f(3)g'(3)+2\times3$. Substitute $f(3)=-1$, $f'(3)=-2$, $g(3)=8$, and $g'(3)=2$ into the above formula: $h'(3)=(-2)\times8+(-1)\times2 + 6=-16-2 + 6=-12$.

Step6: Find the tangent - line equation

The equation of a tangent line is $y - y_{0}=m(x - x_{0})$, where $(x_{0},y_{0})=(3,1)$ and $m = h'(3)=-12$. So $y-1=-12(x - 3)$, which simplifies to $y=-12x+37$.

Answer:

$y=-12x + 37$