Respuesta :
to find the critical points, we need to find the first derivative and equal it to zero as the following :
y = x * tan(x)
y ' = x * sec^2(x) + tan(x) * 1
y ' = ( x / cos^2(x) ) + ( sin(x) / cos(x) ) ---> LCD
y ' = ( x / cos^2(x) ) + ( sin(x)*cos(x) / cos(x)*cos(x) )
y ' = ( x / cos^2(x) ) + ( sin(x)*cos(x) / cos^2(x) )
y ' = ( x + sin(x)*cos(x) ) / cos^2(x)
0 = sec^2(x) * ( x + sin(x) * cos(x) )
0 = sec^2(x) -----> no go
x + sin(x) * cos(x) = 0 -----> can't be solve algebraically, so we are going to use a calculator.
the only solution point is 0, which is the critical point.
to find the interval of concavities, we need to find the points of inflection by finding the second derivative and equal it to zero as the following:
y '' = x * 2sec(x)sec(x)tan(x) + sec^2(x) * 1 + sec^2(x)
y '' = 2x * sec^2(x)tan(x) + 2sec^2(x)
0 = 2x * sec^2(x)tan(x) + 2sec^2(x)
0 = 2sec^2(x) * [ x * tan(x) + 1 ]
0 = sec^2(x) * [ x * tan(x) + 1 ]
0 = sec^2(x) ---> no go ( this will a vertical asymptote ) which is π/2
x * tan(x) + 1 = 0 -----> can't be done algebraically, we are going to use a calculator
solutions are which are the points of inflection:
±18.7964043662102
±15.6441283703330
±12.4864543952238
±9.31786646179107
±6.12125046689807
±2.79838604578389
we need to check before and after each number into the second derived equation, so i will show an example as 2.79838604578389
y '' = 2x * sec^2(x)tan(x) + 2sec^2(x) ------> when x = 2
y '' = 2*2 * sec^2(2)tan(2) + 2sec^2(2)
y '' ≈ -39 ( negative means concaving down )
concave down: ( π/2 , 2.79838604578389 )
we can categorize each number, if you have a limit to the interval, it will be great.
y = x * tan(x)
y ' = x * sec^2(x) + tan(x) * 1
y ' = ( x / cos^2(x) ) + ( sin(x) / cos(x) ) ---> LCD
y ' = ( x / cos^2(x) ) + ( sin(x)*cos(x) / cos(x)*cos(x) )
y ' = ( x / cos^2(x) ) + ( sin(x)*cos(x) / cos^2(x) )
y ' = ( x + sin(x)*cos(x) ) / cos^2(x)
0 = sec^2(x) * ( x + sin(x) * cos(x) )
0 = sec^2(x) -----> no go
x + sin(x) * cos(x) = 0 -----> can't be solve algebraically, so we are going to use a calculator.
the only solution point is 0, which is the critical point.
to find the interval of concavities, we need to find the points of inflection by finding the second derivative and equal it to zero as the following:
y '' = x * 2sec(x)sec(x)tan(x) + sec^2(x) * 1 + sec^2(x)
y '' = 2x * sec^2(x)tan(x) + 2sec^2(x)
0 = 2x * sec^2(x)tan(x) + 2sec^2(x)
0 = 2sec^2(x) * [ x * tan(x) + 1 ]
0 = sec^2(x) * [ x * tan(x) + 1 ]
0 = sec^2(x) ---> no go ( this will a vertical asymptote ) which is π/2
x * tan(x) + 1 = 0 -----> can't be done algebraically, we are going to use a calculator
solutions are which are the points of inflection:
±18.7964043662102
±15.6441283703330
±12.4864543952238
±9.31786646179107
±6.12125046689807
±2.79838604578389
we need to check before and after each number into the second derived equation, so i will show an example as 2.79838604578389
y '' = 2x * sec^2(x)tan(x) + 2sec^2(x) ------> when x = 2
y '' = 2*2 * sec^2(2)tan(2) + 2sec^2(2)
y '' ≈ -39 ( negative means concaving down )
concave down: ( π/2 , 2.79838604578389 )
we can categorize each number, if you have a limit to the interval, it will be great.
