$f(x)=\frac9{x+1}\\ f'(x)=-\frac9{(x+1)^2}\\ f'(x)=-1\ \iff\ -\frac9{(x+1)^2}=-1\ \to \ \frac9{(x+1)^2}=1\ \to \ (x+1)^2=9\\ |x+1|=3\ \to \ x+1=3\ \vee\ x+1=-3\\ x_1=2\ \vee\ x_2=-4\\ f(x_1)=f(2)=\frac9{2+1}=3\\ f(x_2)=f(-4)=\frac9{-4+1}=-3$

First tangent line:

$y=f'(x_1)\cdot (x-x_1)+f(x_1)\ \to \ y=-1(x-2)+3\ \to \ y=-x+5$

Second tangent line:

$y=f'(x_2)\cdot (x-x_2)+f(x_2)\ \to \ y=-1(x+4)-3\ \to \ y=-x-7$

Notice: slope of -1 means that both $f'(x_1), \ f'(x_2)$ are equal to -1, so $f'(x_1)=-1 \ and \ f'(x_2)=-1$

6 0

3 years ago

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So, the additive inverse of -7/8 is

zlopas [31]

Answer:

Im prett sure its 7/8

Step-by-step explanation:

The additive inverse of a number is a number that, when added, goes to zero. For example, 7 is -7 because when added it equals 0 (or 7-7)

8 0

3 years ago