How do I find the numbers #c# that satisfy the Mean Value Theorem for #f(x)=3x^2+2x+5# on the interval #[-1,1]# ?

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Answer 1 · 2018-05-02T14:09:51

# c=0 #

We seek to verify the Mean Value Theorem for the function

# f(x) = 3x^2+2x+5# on the interval #[-1,1]#

The Mean Value Theorem, tells us that if #f(x)# is differentiable on a interval #[a,b]# then #EE \ c in [a,b]# st:

# f'(c) = (f(b)-f(a))/(b-a) #

So, Differentiating wrt #x# we have:

# f'(x) = 6x+2 #

And we seek a value #c in [-1,1]# st: # f'(c) = (f(1)-f(-1))/(1-(-1)) #

# :. 6c+2 = ((3+2+5)-(3-2+5))/(2) #

# :. 6c + 2 = 4/2 #

# :. 6c + 2 = 2 #

# :. 6c = 0 #

# :. c=0 #