- #1

karush

Gold Member

MHB

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View attachment 9236

image to avoid typos

$f'(x)=\dfrac{2}{\left(x+2\right)^2}$

so then at slope $\dfrac{1}{2}$

$\dfrac{2}{\left(x+2\right)^2}=\dfrac{1}{2}$

isolate x

$4=(x+2)^2=x^2+4x+4$

then

$0=x(x+4)$

so

$x=0,-4$

thus the slope is $\dfrac{1}{2}$ at $(0,0), and \left(-4,2\right)\quad (C)$

image to avoid typos

so then at slope $\dfrac{1}{2}$

$\dfrac{2}{\left(x+2\right)^2}=\dfrac{1}{2}$

isolate x

$4=(x+2)^2=x^2+4x+4$

then

$0=x(x+4)$

so

$x=0,-4$

thus the slope is $\dfrac{1}{2}$ at $(0,0), and \left(-4,2\right)\quad (C)$

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