Understanding the Problem

The function g(x) is defined piecewise and is differentiable. For a piecewise function to be differentiable at the point where the definition changes (here at x=3), it must be both continuous and have the same derivative from both sides at that point.

Step 1: Ensure Continuity at x=3

For g(x) to be continuous at x=3, the left-hand limit (from the first piece) must equal the right-hand limit (from the second piece) and must equal the function value at x=3.

Left-hand limit (x→3⁻): $\underset{x\to 3^{-}}{\lim }g\left(x\right)=k\sqrt{x+1}=k\sqrt{3+1}=k\sqrt{4}=2k$

Right-hand limit (x→3⁺): $\underset{x\to 3^{+}}{\lim }g\left(x\right)=mx+2=3m+2$

For continuity, these must be equal:

$2k=3m+2$ (Equation 1)

Step 2: Ensure Differentiability at x=3

For g(x) to be differentiable at x=3, the left-hand derivative must equal the right-hand derivative.

First, find the derivative of each piece.

For 0 ≤ x ≤ 3: $g^{\prime }\left(x\right)=\frac{d}{dx}\left(k\sqrt{x+1}\right)=k·\frac{1}{2\sqrt{x+1}}=\frac{k}{2\sqrt{x+1}}$

Left-hand derivative at x=3: $g^{\prime }\left(3^{-}\right)=\frac{k}{2\sqrt{3+1}}=\frac{k}{2·2}=\frac{k}{4}$

For 3 < x ≤ 5: $g^{\prime }\left(x\right)=\frac{d}{dx}(mx+2)=m$

Right-hand derivative at x=3: $g^{\prime }\left(3^{+}\right)=m$

For differentiability, these must be equal:

$\frac{k}{4}=m$ (Equation 2)

Step 3: Solve the System of Equations

We now have two equations with two variables, k and m.

From Equation 2: $m=\frac{k}{4}$

Substitute this into Equation 1:

$2k=3\left(\frac{k}{4}\right)+2$

$2k=\frac{3k}{4}+2$

Multiply every term by 4 to eliminate the fraction:

$8k=3k+8$

Solve for k:

$8k-3k=8$

$5k=8$

$k=\frac{8}{5}$

Now, find m using Equation 2:

$m=\frac{k}{4}=\frac{1}{4}·\frac{8}{5}=\frac{8}{20}=\frac{2}{5}$

Step 4: Find the Final Answer (k + m)

$k+m=\frac{8}{5}+\frac{2}{5}=\frac{10}{5}=2$

Final Answer

The value of k + m is $2$.

Related Topics & Formulae

Key Concepts

• Continuity: A function f(x) is continuous at a point x = c if: $\underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to c^{+}}{\lim }f\left(x\right)=f\left(c\right)$
• Differentiability: A function is differentiable at a point if its derivative exists there. For a piecewise function, this requires the left-hand and right-hand derivatives to be equal at the boundary point. Differentiability implies continuity.
• Piecewise Functions: Functions defined by different expressions over different intervals. Special attention must be paid to the points where the expression changes.

Important Formulae

• Derivative of a square root: $\frac{d}{dx}\sqrt{u}=\frac{1}{2\sqrt{u}}·\frac{du}{dx}$
• Derivative of a linear function: $\frac{d}{dx}(ax+b)=a$