Find any relative extrema of the function. (round your answers to three decimal places.) f(x) = arcsec(x) − 8x relative maximum (x, y)= relative minimum (x, y) = can someone me with this question?

(x) = arcsec(x) − 8x

f'(x) = d/dx( arcsec(x) − 8x )

 1/xsqrt( x^2 - 1) - 8

f'(x) = 0

1/xsqrt( x^2 - 1) - 8 = 0

8 x sqrt (x^2-1) = 1

 ( 8 x sqrt (x^2-1) )^2 = 1

64 x^2 ( x^2 - 1) = 1

64 x^4 - 64 x^2 =1

64 x^4 - 64 x^2 - 1 = 0

x = 1.00766 , - 1.00766

 x =   - 1.00766

f(- 1.00766) = arcsec(- 1.00766) − 8( - 1.00766)

f( - 1.00766 ) = 11.07949

x = 1.00766

f(1.00766) = arcsec(1.00766) − 8( 1.00766)

f(1.00766 ) = -7.93790

relative maximum (x, y) = (- 1.00766 , 11.07949 ) relative minimum (x, y) = ( 1.00766 , -7.93790  )

60°

step-by-step explanation:

the measure angle of c in an isosceles triangle abc is 60° because in isosceles triangle,the measures of all the angles is 60°.