1. ## Delta Values

I'm stuck on this particular problem. It's given me headaches just staring at it. Can anyone help me with this? Question is:
To 2 decimal places, find the value of $\displaystyle \Delta\;y$ and dy for the given values of $\displaystyle \Delta\;x$ and dx.

$\displaystyle y = 2(x^2 - 3x)^2$
$\displaystyle x = 4$
$\displaystyle \Delta\;x = 0.1$

2. ## Ok

Originally Posted by R3ap3r
I'm stuck on this particular problem. It's given me headaches just staring at it. Can anyone help me with this? Question is:
To 2 decimal places, find the value of $\displaystyle \Delta\;y$ and dy for the given values of $\displaystyle \Delta\;x$ and dx.

$\displaystyle y = 2(x^2 - 3x)^2$
$\displaystyle x = 4$
$\displaystyle \Delta\;x = 0.1$
$\displaystyle \Delta{y}\approx{f'(x)\cdot{\Delta{x}}}$

3. Originally Posted by R3ap3r
I'm stuck on this particular problem. It's given me headaches just staring at it. Can anyone help me with this? Question is:
To 2 decimal places, find the value of $\displaystyle \Delta\;y$ and dy for the given values of $\displaystyle \Delta\;x$ and dx.

$\displaystyle y = 2(x^2 - 3x)^2$
$\displaystyle x = 4$
$\displaystyle \Delta\;x = 0.1$
Remember that
$\displaystyle \Delta x =dx$

The change in y is given by

$\displaystyle \Delta y =f(x+\Delta x) -f(x)=f(4.1)-f(4)=40.68-32=8.68$

$\displaystyle y=f(x) \iff \frac{dy}{dx}=f'(x) \iff dy=f'(x)dx$

$\displaystyle dy=f'(4)(0.1)=(80)(0.1)=8$