Question

The the function's value will always be greater than or equal to the local linear approximation of a function f if, for all x in an interval containing the point of tangency,

f "(x) < 0

f "(x) > 0

f '(x) < 0

f '(x) > 0

Answer 1

@ikram002p I think it's f"(x)>0 since that means the function is concave up and would look like this

Answer 2

@goformit100 @Hero got any ideas?

Answer 3

@Kainui kind of confused about this what do u think ?

Answer 4

This is definitely about the second derivative being interpreted as the average. See, there is actually something called the forward, backwards and middle derivative definitions and here I'll use the middle one and it looks like this: \[\Large f'(x)=\lim_{h \rightarrow 0} \frac{f(x+\frac{h}{2})-f(x-\frac{h}{2})}{h}\]

Now when we take the second derivative we have:

\[\Large f''(x)=\lim_{h \rightarrow 0} \frac{1}{h} (\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h})\] I'll continue deriving how the second derivative is really the average in a moment.

Answer 5

\[\Large f''(x)=\lim_{h \rightarrow 0} \frac{1}{h^2} (f(x+h)-2f(x)+f(x-h))\] \[\Large f''(x)=\lim_{h \rightarrow 0} \frac{1}{2h^2} [\frac{f(x+h)+f(x-h)}{2}-f(x)]\]

Now it should be clear that f(x+h) and f(x-h) are the points next to f(x). So if we add them together and divide by 2, this is the average of the neighboring points (this is also why the Laplacian is commonly found in differential equations and physics)

So notice, if the neighboring points is higher than the function's value at that point, the second derivative is positive.

If the average of the neghboring points is lower than the point, the second derivative will be negative.

Answer 6

If you need help understanding I can answer all your questions.

Answer 7

I want to know you went to get that second derivative formula from the first one

Answer 8

But would I be correct in my simpler answer that the answer is f "(x) > 0
since the graph will be concave up like in this pic