function

Let

f(x) = 2x^2 - 3 if x <= 2

gx) = ax - 7 if x > 2
Find a if the graph of y = f(x) is continuous

Hello Guest!

$\color{BrickRed}f(x)=2x^2-3\\ \color{blue}f'(x)=4x=a\\ \color{BrickRed}g(x)=ax-7\\ 4x = a\ inserted\ in\ g (x)\\ g(x)=4x^2-7\\ \color{blue} f(x)=g(x)$

$2x^2-3=4x^2-7\\ 2x^2=4\\ \color{blue}x=\pm \sqrt{2}$

$y=2\cdot 2-3$

$y=1$

$g(x)=ax-7\\ 1=a\cdot \sqrt{2}-7\\ a=4\cdot \sqrt{2}$

$a=5.65685\\ g(x)=5.65685x-7$

\(y\in \{f(x)=2x^2-3\}\ |-∞  LaTeX is crazy today!

y $\in$ {f(x)=$2x^2$-3} | ($-∞$ < x $\le \sqrt{2}$ )

y $\in$ {g(x)=5.65685y-7} | ($\sqrt{2}$ < x $\le 2$)

y $\in$ {g(x)=5.65685x-7}    | ($2$ < x < $∞$)

  !

Thanks Alan!

If corners and crossings are allowed in the line, it goes like this:

$f(x) = 2x^2 - 3\ |\ if\ x <= 2\\ g(x) = ax - 7\ |\ if\ x > 2$

$x=2$

$f(2)=g(2)\\ 2\cdot 2^2-3=2a-7\\ a=\large \frac{2\cdot 2^2-3+7}{2}$

$a=6$

$g(x)=6x-7$

  !