Find the values of a and b such that $h(x)=\begin{cases}ax^2+7x & x<1\\ 2x+b & x\ge 1\end{cases}$ differentiable at $x=1$

Key concept: For a piecewise function to be differentiable at a point, it must first be continuous there, and its left and right derivatives must also match. That gives two equations for $a$ and $b$.

Step by step

Step 1: Set the derivatives equal at $x=1$

For $x<1$,

$h(x)=ax^2+7x \Rightarrow h'(x)=2ax+7$

For $x\ge 1$,

$h(x)=2x+b \Rightarrow h'(x)=2$

Differentiability at $x=1$ requires

$2a(1)+7=2$

so

$2a+7=2$

Step 2: Use continuity at $x=1$

The two pieces must also have the same value at $x=1$:

Left side: $a(1)^2+7(1)=a+7$

Right side: $2(1)+b=2+b$

So

$a+7=2+b$

Step 3: Solve the system

From $2a+7=2$,

$2a=-5 \Rightarrow a=-\frac{5}{2}$

Then substitute into $a+7=2+b$:

$-\frac{5}{2}+7=2+b$

$\frac{9}{2}=2+b$

$b=\frac{5}{2}$

Step 4: Find the sum

$a+b=-\frac{5}{2}+\frac{5}{2}=0$

Final answer

$0$

Pitfall alert

A common mistake is using only continuity or only derivative matching. Differentiability at a junction point requires both conditions. Also, make sure the derivative of $ax^2$ is $2ax$, not $a\cdot 2x$ without the coefficient handled correctly.

Try different conditions

If the question asked only for continuity at $x=1$, you would use just $a+7=2+b$. If it asked only for differentiability, you would still need the derivative-matching equation together with continuity to determine both constants.

Further reading

piecewise function, continuity, differentiability