Delta Values

**R3ap3r** · April 10th 2008, 02:06 PM

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 $\Delta\;y$ and dy for the given values of $\Delta\;x$ and dx.

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

**Mathstud28** · April 10th 2008, 02:10 PM

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 $\Delta\;y$ and dy for the given values of $\Delta\;x$ and dx.

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

$\Delta{y}\approx{f'(x)\cdot{\Delta{x}}}$

**TheEmptySet** · April 10th 2008, 02:13 PM

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 $\Delta\;y$ and dy for the given values of $\Delta\;x$ and dx.

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

Remember that
$\Delta x =dx$

The change in y is given by

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

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

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

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