Question

Let $g$ be a real-valued differentiable function on $R$ such that $g(x) = 3e^{x-2} + 4\int_{2}^{x} \sqrt{2t^2 + 6t + 5}dt$ $\forall x \in R$ and let $g^{-1}$ be the inverse function of $g$. If $(g^{-1})'(3)$ is equal to $p/q$ where $p$ and $q$ are relatively prime natural numbers, then find $(p+q)$.

• A. 24
• B. 34
• C. 44
• D. 64

Answer 1

Answer: (A)

24

Answer 2

To find $(g^{-1})'(3)$, we use the formula for the derivative of an inverse function: $(g^{-1})'(y) = \frac{1}{g'(x)}$ where $y = g(x)$.

1. Find $x$ such that $g(x) = 3$. The given function is $g(x) = 3e^{x-2} + 4\int_{2}^{x} \sqrt{2t^2 + 6t + 5}dt$. Let's test $x=2$: $g(2) = 3e^{2-2} + 4\int_{2}^{2} \sqrt{2t^2 + 6t + 5}dt$ $g(2) = 3e^0 + 4 \cdot 0$ $g(2) = 3 \cdot 1 + 0$ $g(2) = 3$ So, when $g(x) = 3$, we have $x = 2$. Therefore, we need to calculate $(g^{-1})'(3) = \frac{1}{g'(2)}$.

2. Find $g'(x)$. Differentiate $g(x)$ with respect to $x$: $g'(x) = \frac{d}{dx}\left(3e^{x-2}\right) + \frac{d}{dx}\left(4\int_{2}^{x} \sqrt{2t^2 + 6t + 5}dt\right)$ Using the chain rule for the exponential term and the Fundamental Theorem of Calculus for the integral term: $\frac{d}{dx}(3e^{x-2}) = 3e^{x-2} \cdot \frac{d}{dx}(x-2) = 3e^{x-2} \cdot 1 = 3e^{x-2}$. $\frac{d}{dx}\left(4\int_{2}^{x} \sqrt{2t^2 + 6t + 5}dt\right) = 4\sqrt{2x^2 + 6x + 5}$. So, $g'(x) = 3e^{x-2} + 4\sqrt{2x^2 + 6x + 5}$.

3. Find $g'(2)$. Substitute $x=2$ into the expression for $g'(x)$: $g'(2) = 3e^{2-2} + 4\sqrt{2(2)^2 + 6(2) + 5}$ $g'(2) = 3e^0 + 4\sqrt{2(4) + 12 + 5}$ $g'(2) = 3 \cdot 1 + 4\sqrt{8 + 12 + 5}$ $g'(2) = 3 + 4\sqrt{25}$ $g'(2) = 3 + 4 \cdot 5$ $g'(2) = 3 + 20$ $g'(2) = 23$.

4. Calculate $(g^{-1})'(3)$. $(g^{-1})'(3) = \frac{1}{g'(2)} = \frac{1}{23}$.

5. Determine $(p+q)$. The problem states that $(g^{-1})'(3) = p/q$, where $p$ and $q$ are relatively prime natural numbers. We have $\frac{p}{q} = \frac{1}{23}$. Thus, $p=1$ and $q=23$. These are relatively prime natural numbers. We need to find $(p+q)$: $p+q = 1 + 23 = 24$.

The final answer is $\boxed{\text{24}}$.

Explanation of the solution:

1. Identify $x_0$ such that $g(x_0)=3$. By inspection, $g(2)=3e^{2-2} + 4\int_{2}^{2} \sqrt{2t^2 + 6t + 5}dt = 3e^0 + 0 = 3$. So $x_0=2$.
2. The derivative of the inverse function is given by $(g^{-1})'(3) = \frac{1}{g'(2)}$.
3. Differentiate $g(x)$ using the chain rule and the Fundamental Theorem of Calculus: $g'(x) = \frac{d}{dx}(3e^{x-2}) + \frac{d}{dx}(4\int_{2}^{x} \sqrt{2t^2 + 6t + 5}dt) = 3e^{x-2} + 4\sqrt{2x^2 + 6x + 5}$.
4. Evaluate $g'(2)$: $g'(2) = 3e^{2-2} + 4\sqrt{2(2)^2 + 6(2) + 5} = 3e^0 + 4\sqrt{8+12+5} = 3 + 4\sqrt{25} = 3 + 4(5) = 3+20 = 23$.
5. Calculate $(g^{-1})'(3) = \frac{1}{23}$.
6. Given $(g^{-1})'(3) = p/q$, we have $p=1$ and $q=23$.
7. Find $p+q = 1+23 = 24$.

The final answer is $\boxed{\text{24}}$.