Question

verify lagrange's mean value theorem for f(x) =x^3-x^2-5x+3 in[0,4]​

Answer 1

To Verify that;

Lagrange's mean value theorem for f(x) =x^3-x^2-5x+3 in [0,4]​

Proof;

The given function is  f(x) = $x^{3} - x^{2} -5x + 3$

f is a polynomial function, so it is continuous in [0,4] and is differentiable in (0,4) whose derivative is $3x^{2} - 2x - 5$.

Now, f(0) = 3 and f(4) = 64 - 16 - 20 + 3 = 31

Therefore, $\frac{f(b) - f(a)}{b - a}$ = $\frac{31 - 3}{4 - 0}$ = 7.

Mean Value Theorem states that there exists a point C ∈(0,4) such that $f^{'} (c)$ = 7

So, $3c^{2} - 2c - 5$ = 7

  $3c^{2} - 2c - 12$ = 0

c = $\frac{ - 1 + \sqrt{37} }{3}$ and $\frac{ -1 - \sqrt{37} }{3}$

c = 1.69  and c = - 2.36

Thus, C ∈(0,4).

Hence,  Lagrange's mean value theorem is verified.

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Answer 2

Answer: lagrange's mean value theorem for f(x) =x^3-x^2-5x+3 in[0,4]​ is verified in this interval

Step-by-step explanation:

To Verify that;

Lagrange's mean value theorem for f(x) =x^3-x^2-5x+3 in [0,4]​

Proof;

The given function is  f(x) =

f is a polynomial function, so it is continuous in [0,4] and is differentiable in (0,4) whose derivative is :-

Now, f(0) = 3 and f(4) = 64 - 16 - 20 + 3 = 3

Therefore,  =  = 7.

Mean Value Theorem states that there exists a point C ∈(0,4) such that  = 7

c = 1.69  and c = - 2.36

Thus, C ∈(0,4).

the lagrange theoram states that the line joining the end points it will be a tangent at some points where the tangent is parellel to the line joining these end points

Hence,  Lagrange's mean value theorem is verified.

Thus verified and prove that lagrange's mean value theorem for f(x) =x^3-x^2-5x+3 in[0,4]​ is satisfied in this interval.

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