Local max/min

gymnast24

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Feb 26, 2006
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Consider the function f(x), whose derivative f'=0 at x=2 and whose second derivative f''(x)= x^2+1. Choose one

a. f has a local maximum at x=2
b. f has a local minimum at x=2
c. f has a point of inflection at x=2
d. f has no critical numbers

So far I've set the second derivative to zero and got nothing so now I don't know what to do! I dont' get how the first derivative can be zero and the second derivative of the same function is x^2+1
 
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Hello, gymnast24!

Consider the function \(\displaystyle f(x)\), whose derivative \(\displaystyle f'\,=\,0\) at \(\displaystyle x\,=\,2\)
and whose second derivative is: \(\displaystyle f''(x)\:=\:x^2\,+\,1\)

a. \(\displaystyle f\) has a local maximum at \(\displaystyle x\,=\,2\,\)
b. \(\displaystyle f\) has a local minimum at \(\displaystyle x\,=\,2\)
c. \(\displaystyle f\) has a point of inflection at \(\displaystyle x\,=\,2\)
d. \(\displaystyle f\) has no critical numbers


So far I've set the second derivative to zero . . . why?
I dont' get how the first derivative can be zero . . .they plugged in x = 2
and the second derivative of the same function is x^2+1
They told us that the first derivative is zero when \(\displaystyle x\,=\,2\).
\(\displaystyle \;\;\)So there is a critical point there . . . a horizontal tangent.

It could be a maximum, a minimum, an inflection point, or none of these.
\(\displaystyle \;\;\)How do we find out? . . . the second derivative test.

Plug \(\displaystyle x\,=\,2\) into the second derivative: \(\displaystyle \,f''(2) \:=\:2^2\,+\,1\:=\:5\)
\(\displaystyle \;\;\)Since the second derivative is positive, the graph is concave up there: \(\displaystyle \cup\)

Therefore, there is a minimum at \(\displaystyle x\,=\,2\;\;\) . . . answer (b)
 
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There is no better way to kill learning than just to supply an answer.
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It just as simple as the old tale “teach to fish or give a fish”.
 
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