Given that f(1)=2 and f'(1)=4, find the derivative of f(x)loge(x) when x=1
1 answer:
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Answer:
maximum: y = 1
minimum: y = 0.
Step-by-step explanation:
Here we have the function:
y = f(x) = √(1 + x^2 - 2x)
we want to find the minimum and maximum in the segment [0, 1]
First, we evaluate in the endpoints, which are 0 and 1.
f(0) =√(1 + 0^2 - 2*0) = 1
f(1) = √(1 + 1^2 - 2*1) = 0
Now let's look at the critical points (the zeros of the first derivate)
To derivate our function, we can use the chain rule:
f(x) = h(g(x))
then
f'(x) = h'(g(x))*g(x)
Here we can define:
h(x) = √x
g(x) = 1 + x^2 - 2x
Then:
f(x) = h(g(x))
f'(x) = 1/2*( 1 + x^2 - 2x)*(2x - 2)
f'(x) = (1 + x^2 - 2x)*(x - 1)
f'(x) = x^3 - 3x^2 + x - 1
this function does not have any zero in the segment [0, 1] (you can look it in the image below)
Thus, the function does not have critical points in the segment.
Then the maximum and minimum are given by the endpoints.
The maximum is 1 (when x = 0)
the minimum is 0 (when x = 1)
