derivative function

[wendywoo]

suppose that f(0)=3, and f' is the function shown below. Let g(x)= (x^2+1)f(x)
Evaluate g'(0).
i took the derivative of g(x) and plugged in the zero in for x.
i got g'(0)= (1)f'(0)+f(0)(0)=0
0 is my final answer but i think the correct answer is supposed to be 1. why is that? The derivative of 0 should be 0 yes? help!!

 

[mmm4444bot]

f' is the function shown below

We can't see it.



g'(0) = (1)f'(0) + f(0)(0)

Hence, g`(0) = f`(0)

Does the graph of f`(x) pass through the origin ?

 

[wendywoo]

the graph passes through the point (0,1) but i don't understand why the derivative of f'(0) is 1.

 

[mmm4444bot]

i don't understand why the derivative of f'(0) is 1

Be careful with your terminology. We're not talking about the derivative of f`(0).

We're talking about the derivative of some unknown function f(x) at the point where x = 0.

We do not have a graph of f(x), and we do not need one.

To evaluate g`(0) from the given information, you need to know two values: f(0) and f`(0).

The exercise told you the value of f(0) directly, and they gave you a graph to look up the value of f`(0).

The given graph tells us that f`(0) = 1 because it IS the graph of y = f`(x).

Since that graph passes through the point (0, 1), we have y = 1 when x = 0.

In other words: f`(0) = 1

You already showed that g`(0) = f`(0), so by the Transitive Property of Equality g`(0) = 1, too.