Question

The function f'(x) = cos(ln x) is the first derivative of a twice differentiable function, f(x). On the interval 0

Answer 1

x =
`e^{(\pi)/(2)}`

Explanation:

Data provided in the question:

f'(x) = cos(ln x)

For finding points of maxima or minima , put f'(x) = 0

therefore,

cos(ln x) = 0

or

cos(ln x) =
`\cos((\pi)/(2))`

or

ln x =
`(\pi)/(2)`

or

x =
`e^{(\pi)/(2)}`

now to check for maxima or minima at x =
`e^{(\pi)/(2)}`

f"(x) =
`-\sin(\ln x)*(1)/(x)`

thus

at x =
`e^{(\pi)/(2)}`

f"(x) =
`-\sin(\ln (e^{(\pi)/(2)}))*\frac{1}{e^{(\pi)/(2)}}`

=
`-\sin(e^{(\pi)/(2)})* e^{-(\pi)/(2)}}`

=
`-1* e^{-(\pi)/(2)}}` < 0

hence maxima at x =
`e^{(\pi)/(2)}`