Find a and b such that the function
$$f(x)= x^2-4x+4$$ if $$x \le 3$$
f(x) =$$ax+b$$ if $$x 3$$
is differentiable everywhere.

Question

Find a and b such that the function
$$f(x)= x^2-4x+4$$ if $$x \le 3$$
f(x) =$$ax+b$$ if $$x > 3$$
is differentiable everywhere.

solomonzelman · Answer

$\color{#000000}{f(x)=\begin{cases}  x^2-4x+4,~~~ x\le 3 \ ax+b,~~~ x >3 \end{cases}  }$

like this?

solomonzelman · Answer

Obviously, at first, you need f(3)=f(3) for both parts of the function, to make the function continous.

solomonzelman · Answer

Then, you have to set f'(3)=f'(3), to make the derivative equivalent so that you have the same slope (otherwise the function wouldn't be differentiable at x=3)

solomonzelman · Answer

All you need to do is:

Find f(3) for both parts of $f(x)$, and set
them equivalent to each other.

Differentiate both parts of the function
and evaluate them at x=3.

This will give you the 1st and 2nd equations to solve, respectively.

solomonzelman · Answer

A similar example. I used a different
function, but the method is the same.
$\color{#000080}{\text{---------------------------------------------}  }$
Find $\color{#000000}{a  }$ and $\color{#000000}{b  }$ such that the function,
$\color{#000000}{f(x)=\begin{cases}  2x^2-10x-15,~~~ x\le 5 \ bx+5a,~~~ x >5 \end{cases}  }$
is differentiable at every points.
$\color{#000080}{\text{---------------------------------------------}  }$
In order for $\color{#000000}{f(x)  }$ to be differentiable,
leastwise it must be continuous, and
for $\color{#000000}{f(x)  }$ to be continuous, the parts
of $\color{#000000}{f(x)  }$ should have the same point
at $\color{#000000}{x=5  }$.

$\color{#000000}{f_1(x)=2x^2-10x-15  }$
$\color{#000000}{f_1(\color{red}{5})=2\color{red}{(5)}^2-10\color{red}{(5)}-15=-15  }$

$\color{#000000}{f_2(x)=bx+5a  }$
$\color{#000000}{f_2(\color{red}{5})=b\color{red}{(5)}+5a=5b+5a   }$

The function is obviously continuous
at every point asides from $\color{#000000}{x=5  }$, as
well as the fact that it is true that it is
differentiable at every other point.

The only thing to be verified (to solve
$\color{#000000}{a  }$ and $\color{#000000}{b  }$) is that $\color{#000000}{f(x)  }$ is continuous and
differentiable at $\color{#000000}{x=5  }$. For continuity
at $\color{#000000}{x=5  }$ I set $\color{#000000}{f_1(5)=f_2(5)  }$.

$\color{#000000}{f_1(5)=f_2(5)  }$
$\color{#000000}{-15=5b+5a  }$		(the 1st equation)

Next, I want to make sure that $\color{#000000}{f(x)  }$ is
differentiable at $\color{#000000}{x=5  }$, and to do this
I will differentiate each part of $\color{#000000}{f(x)  }$,
and set $\color{#000000}{f_{1}{}'(5) =f_{2}{}'(5) }$.

$\color{#000000}{f_1(x)=2x^2-10x-15  }$
$\color{#000000}{f_{1}{}'(x) =4x-10 }$
$\color{#000000}{f_{1}{}'(\color{blue}{5}) =4\color{blue}{(5)}-10=10 }$

$\color{#000000}{f_2(x)=bx+5a  }$
$\color{#000000}{f_{2}{}'(x) =b }$
$\color{#000000}{f_{2}{}'(\color{blue}{5}) =b }$

$\color{#000000}{f_{1}{}'(5) =f_{2}{}'(5) }$
$\color{#000000}{10 =b }$		(the 2nd equation)

So, there you have the answer:
$\color{#000000}{10 =b }$
$\color{#000000}{-15=5b+5a  }$
$\color{#000000}{-15=5(10)+5a ~~~\Rightarrow ~~~  a=-13}$

$\Large{\bbox[5pt, #ffffcc ,border:2px solid  black  ]{  \color{#ffffcc}{\frac{\color{red}{a~=~-13}\frac{~}{~}  }{\color{red}{b~=~10\tiny{~~~~~~~~~~~~~~~}~} }}   }}$

anonymous · Answer

Thank you so much for the explanation! @SolomanZelman