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Let

 

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

f(x) = ax - 7 if x > 2


Find a if the graph of y = f(x) is continuous (which means the graph can be drawn without lifting your pencil from the paper)

 Oct 20, 2020
 #1
avatar+10583 
+1

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 < \(∞\))

laugh  !

 Oct 20, 2020
edited by asinus  Oct 20, 2020
edited by asinus  Oct 20, 2020
edited by asinus  Oct 20, 2020
edited by asinus  Oct 20, 2020
edited by asinus  Oct 20, 2020
 #3
avatar+10583 
+1

Thanks Alan!

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

 

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

 

\(x=2\)

 

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

\(a=6\)

 

\(g(x)=6x-7\)

laugh  !

 Oct 20, 2020

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