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I need to find the mean slope of $f(x) = 2x^3-6x^2-90x+6$ on the interval $[-5,8]$. I am getting $\frac{518}{13}$ but this is wrong. Here is my work.

Using the mean value theorem $f'(c)=\frac{f(b) - f(a)}{b - a}$

$f(b) = 2(-5)^3-6(-5)^2-90(-5)+6 = -250-150+450+6 = 56$

$f(a) = 2(8)^3-6(8)^2-90(8)+6 = 1024-384-66 = 574$

$\frac{56-574}{-5-8}=\frac{518}{13}$

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  • $\begingroup$ The mean slope is just the slope of the secant line. $\endgroup$ – coreyman317 Mar 31 at 20:57
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Your error is a simple mistake in computing $f(8)$. To be precise, you write that $$2(8)^3-6(8)^2-90(8)+6 = 1024-384-66,$$ where indeed $2\times8^3=1024$ and $6\times8^2=384$. You seem to have computed $9\times8$ instead of $90\times8$.

Apart from this, your work is entirely correct.

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    $\begingroup$ Yes, but you compute $9\times8$ instead of $90\times8$ (I have updated my answer to point this out more clearly) $\endgroup$ – Servaes Mar 31 at 21:02
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    $\begingroup$ my mistake! sorry! $\endgroup$ – LuminousNutria Mar 31 at 21:02
  • $\begingroup$ Happens to the best of us ;) $\endgroup$ – Servaes Mar 31 at 21:03

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