SL 2.4—Key features of graphs, intersections using technology

We are given $f^{\prime }(x)=6x^2+3e^x$ and the point $(0,-2)$. We need to find $f(x)$ by integrating and applying the initial condition.

$f(x)=\int (6x^2+3e^x)dx=2x^3+3e^x+C$

Substituting $(0,-2)$: $-2=2(0{)}^3+3e^0+C=0+3+C$ So $C=-5$.

Therefore $f(x)=2x^3+3e^x-5$, which matches the first option.

Each option must satisfy two conditions: its derivative must equal $6x^2+3e^x$, and $f(0)=-2$.

The derivative of $12x+3e^x-2$ is $12+3e^x$, which is not $6x^2+3e^x$. Eliminated.

The derivative of $2x^3-3e^x+1$ is $6x^2-3e^x$, which has the wrong sign on the $e^x$ term. Eliminated.

This has the correct derivative $6x^2+3e^x$, but $f(0)=0+3+2=5\ne -2$. Eliminated.

Check: derivative is $6x^2+3e^x$ ✓, and $f(0)=0+3-5=-2$ ✓. Correct answer.

We are told $T(0)=0$. Substituting: $0=a\ln (0+b)=a\ln (b)$

Since $a>0$, we require $\ln (b)=0$, which gives $b=1$.

With $b=1$: $4=a\ln (9+1)=a\ln (10)$ So $a=\frac{4}{\ln (10)}\approx 1.74$.

The value of $b=1$ ensures $T(0)=0$ is satisfied, which is the defining condition here.

At $t=0$, $T=0$, so $a\ln (b)=0$. Since $a>0$, we need $\ln (b)=0$, meaning $b=1$.

$\ln (3)\approx 1.099\ne 0$, so $T(0)=a\ln (3)\ne 0$. Eliminated.

$\ln (9)\approx 2.197\ne 0$, so $T(0)\ne 0$. Eliminated.

$\ln (e)=1\ne 0$, so $T(0)=a\ne 0$. Eliminated.

$\ln (1)=0$, so $T(0)=a\cdot 0=0$ ✓. The answer is $b=1$.