The value of c in the Lagrange's mean value theorem for the function
ƒ(x) = x3
- 4x2
+ 8x + 11,
when x $\in$ [0, 1] is:
Solution
ƒ(x) = x<sup>3</sup>
- 4x<sup>2</sup>
+ 8x + 11
<br><br>f(0) = 11
<br><br>f(1) = 16
<br><br>Using LMVT
<br><br>f'(c) = ${{f\left( 1 \right) - f\left( 0 \right)} \over {1 - 0}}$
<br><br>$\Rightarrow$ 3c<sup>2</sup>
– 8c + 8 = ${{16 - 11} \over {1 - 0}}$
<br><br>$\Rightarrow$ 3c<sup>2</sup>
– 8c + 3 = 0
<br><br>$\therefore$ c = ${{8 \pm 2\sqrt 7 } \over 6}$
<br><br>$\therefore$ c = ${{4 - \sqrt 7 } \over 3}$ as c $\in$ [0, 1]
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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