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1 answer:
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Answer:
Step-by-step explanation:
The average value theorem sets:
if f (x) is continuous in [a, b] and derivable in (a, b) there is a c Є (a, b) such that
, where
f(a)=f(π/2)=-4*sin(π/2) = -4*1= -4
f(b)=(3π/2)=-4*sin(3π/2) = -4*-1 = 4
⇒
c≅130
<span>Determine the values of x on which the function f(x)=2x^2-x-15/4x^2-12x is discontinuous and verify the type of discontinuity at each point.
A.There is a vertical asymptote at 0 and a hole at 3.
B.There are vertical asymptotes at -5/2 & 0 and a hole at 3.
C.There is a vertical asymptote at 3 and a hole at 0.
D.There are vertical asymptotes at 0 & 3.
</span>
Result:
(-7/4)x^2-13x
The roots are:
x= -52/7
x=0
See attached pictures.
A.There is a vertical asymptote at 0 and a hole at 3.
B.There are vertical asymptotes at -5/2 & 0 and a hole at 3.
C.There is a vertical asymptote at 3 and a hole at 0.
D.There are vertical asymptotes at 0 & 3.
</span>
Result:
(-7/4)x^2-13x
The roots are:
x= -52/7
x=0
See attached pictures.