b.f(x)=x^3=x-4
Typo, makes no sense
c. f(2)=8
f(-1)=-1
so
9/3 average slope =3
f' =3x^2
3 x^2 =3
x= sqrt(1) which is 1 !!!!
but also -1
d f(4) = 2
f(0) = 0
so slope = 2/4 = 1/2
f' = (1/2)x^-1/2 = 1/2
x = 1 again
Is this the correct answers for these questions
Verify the means value theorem holds on the interval shown. Then, find the value c such that f'(c)=(f(b)-f(a))/(b-a)
b.f(x)=x^3=x-4 on [-2,3] c= square root 7/3
c. f(x)= x^3 on [-1,2] c= square root 1
d. f(x)= Sqr. root of x on[0,4]c= square root 1
2 answers
I'll do (b) to show the way. Of course, you probably already know the way...
Assuming you meant
f(x) = x^3+x-4,
f(-2) = -14
f(3) = 26
slope of secant is 40/5 = 8
f' = 3x^2+1
3x^2+1 = 8
3x^2 = 7
...
clearly, √(7/3) is in [-2,3]
Assuming you meant
f(x) = x^3+x-4,
f(-2) = -14
f(3) = 26
slope of secant is 40/5 = 8
f' = 3x^2+1
3x^2+1 = 8
3x^2 = 7
...
clearly, √(7/3) is in [-2,3]
1 answer