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Determine where the function is not differentiable:
Then `f'(x)` fails to exist for `x<=-5` .(x<-5 has only nonreal solutions, while x=-5 causes a division by zero -- a vertical tangent line in this case.)
You could also reason that the derivative of `y=sqrt(x)` only exists for x>0, and `y=sqrt(x+5)` is just the square root function shifted horizontally 5 units to the left.
Note that f(x)=x+1 for x>2,3 for x=2, and -x+5 for x<2
So `f'(x)=1"for" x>2,f'(x)=-1"for" x>2`
Since the left-hand and right-hand derivatives do not agree at 2, the function is not differentiable at 2. Graphically, there is a corner there.
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