Question

$f(x) = \begin{cases} -\frac{2}{3}x - 8 & \text{if } x \le 1\\ \frac{1}{4}x - 6 & \text{if } x > 1 \end{cases}$

          $f(x) = \begin{cases} -\frac{2}{3}x - 8 & \text{if } x \le 1\\ \frac{1}{4}x - 6 & \text{if } x > 1 \end{cases}$
        
f(x) =  -(2)/(3)x - 8    if  x ≤ 1
(1)/(4)x - 6    if  x > 1

Added by Alvaro S.

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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f(x)={(-(2)/(3)x-8 if x<=1),((1)/(4)x-6 if x>1):}
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Transcript

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00:01 Okay, so we want to prove if f is greater than or equal to 1, then the log of x is less than or equal to x minus 1.
00:08 So then what we're going to do is subtract log x from both sides.
00:15 So we need 0 less than or equal to x minus 1 minus the log of x.
00:22 In other words, that the thing on the right hand side of this is always positive, or 0.
00:33 If we put in x equals 1, then the logarithm is 0 and the first part of the term is also going to be 0, and so the answer is 0.
00:45 So the equal sign does occur.
00:48 Okay, so let's take the derivative of the right hand side.
00:53 Let's call that f of x.
00:57 So f prime of x is equal to 1 minus 1 over x...
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