Find Derivative: Solve G'(3) Problem

[Teh]

View attachment 6132

Does it want me to plug in G'(3) into the equation?

 

[MarkFL]

We are told:

\(\displaystyle G(x)=6f(x)-g(x)\)

And so we know:

\(\displaystyle G'(x)=6f'(x)-g'(x)\)

which means:

\(\displaystyle G'(3)=6f'(3)-g'(3)\)

So, what you need to do is determine from the graph the values of $f'(3)$ and $g'(3)$, and plug them into the above...:D

 

[Teh]

We are told:

\(\displaystyle G(x)=6f(x)-g(x)\)

And so we know:

\(\displaystyle G'(x)=6f'(x)-g'(x)\)

which means:

\(\displaystyle G'(3)=6f'(3)-g'(3)\)

So, what you need to do is determine from the graph the values of $f'(3)$ and $g'(3)$, and plug them into the above...:D

will it be -26 when G'(3)

 

[MarkFL]

will it be -26 when G'(3)

What values do you get from the graph for $f'(3)$ and $g'(3)$?

 

[Teh]

What values do you get from the graph for $f'(3)$ and $g'(3)$?

you get f'(3) = 6 and g'(3) = 1

 

[MarkFL]

Around $x=3$, observe that for every increase of 1 unit in $x$, we find $f$ decreases by 3 units...thus $f'(3)=-3$, and I agree that $g'(3)=1$.

 

[Teh]

Around $x=3$, observe that for every increase of 1 unit in $x$, we find $f$ decreases by 3 units...thus $f'(3)=-3$, and I agree that $g'(3)=1$.

how did you get -3 from f'(3)?

 

[MarkFL]

how did you get -3 from f'(3)?

Rise over run...rise is -3 when run is 1...:D