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Verify Mean Value Theorem, if f (x) equals x raised to 3 minus 5x raised to 2 minus 3x in the interval [a, b], where a equals 1 and b equals 3. Find all c (1, 3) for which f (c) = 0.

Q5  Verify Mean Value Theorem, iff (x) = x^3 - 5x^2- 3xin the interval [a, b], where
       a = 1 and b = 3. Find all c \epsilon (1,3) for which f '(c) = 0.

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Condition for M.V.T.
If f ; [ a ,b] \rightarrow R
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then,  their exist a c  in (a,b) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
It is given that
 f (x) = x^3 - 5x^2- 3x  and interval is [1,3]
Now, f being a polynomial function , f (x) = x^3 - 5x^2- 3x is continuous in[1,3] and  differentiable in (1,3) 
And
f(1)= 1^3-5(1)^2-3(1)= 1-5-3=1-8=-7
and
f(3)= 3^3-5(3)^2-3(3)= 27-5.9-9= 18-45=-27
Then, by Mean value theorem we can say that their exist a c  in (1,4) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
f^{'}(c) = \frac{f(3)-f(1)}{3-1}\\ f^{'}(c)= \frac{-27-(-7)}{2}\\ f^{'}(c)= \frac{-20}{2}\\ f^{'}(c)= -10
Now,
f^{'}(x) =3x^2-10x-3\\ f^{'}(c)=3c^2-10c-3\\ -10=3c^2-10c-3\\ 3c^2-10c+7=0\\ 3c^2-3c-7c+7=0\\ (c-1)(3c-7)=0\\ c = 1 \ \ \ and \ \ \ c = \frac{7}{3}
And c=1,\frac{7}{3} \ and \ \frac{7}{3}\ \epsilon \ (1,3)
Hence, mean value theorem is varified for following function f (x) = x^3 - 5x^2- 3x and  c=\frac{7}{3} is the only point where f '(c) = 0

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