Calculus problem #63971

Calculus Level 1

Suppose $f$ is a function defined on the closed interval $-3 \le x \le 4$ with $f(0)=42$ such that the graph of $f',$ the derivative of $f,$ on the interval is as shown in the above diagram. Find the $x$ -coordinates of the points of inflection of $f.$

x=-3, x=4

x=-3, x=2

x=0, x=2

x=-2, x=2

1 solution

Rishabh Deep Singh
Jan 25, 2016

for point of inflection $f^{ ' }\left( x \right)$ =0
The point at which the 2nd derivative switches sign is called the Inflation point. => The point x = 0 is also an inflation point. The slop of f' changes from -ve to +ve. similarly goes for x == 2.
We cannot find 2nd derivative of f at x == 0 because the LHL and RHL are not the same.

I guess you mean f''(x) =0? And f'(x) has a min or max

Peter van der Linden - 4 years, 8 months ago

Yes Exactly

Rishabh Deep Singh - 4 years, 8 months ago

I don't understand why on x=0 the second derivative it´s = 0 Can anyone explain please?

Vicente Cordova - 11 months, 1 week ago

The point at which the 2nd derivative switches sign is called the Inflation point. => The point x = 0 is also an inflation point. The slop of f' changes from -ve to +ve. similarly goes for x == 2. We cannot find 2nd derivative of f at x == 0 because the LHL and RHL are not the same.

Rishabh Deep Singh - 11 months ago

Makes sense. Thank you for your time and answer.

Vicente Cordova - 9 months, 3 weeks ago

I have updated the answer please check.

Rishabh Deep Singh - 11 months ago

Calculus problem #63971

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