Question
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
f(x) = x(x + 4)2 on [0,4]
f(x) = x(x + 4)2 on [0,4]
f(x) = x(x + 4)2
⇒ f(x) = x(x2 + 16 + 8x)
⇒ f(x) = x3 + 8x2 + 16x
Since f(x) is a polynomial function which is everywhere continuous and differentiable.
Therefore, f(x) is continuous on [0, 4] and derivable on (0, 4)
Thus, both the conditions of Lagrange's theorem is satisfied.
Consequently, there exists some $\text{c}\in(0,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}=\frac{\text{f}(4)-\text{f}(0)}{4}$
Now, f(x) = x3 + 8x2 + 16x
⇒ f(x) = 3x2 + 16x + 16,
⇒ f(4) = 64+ 128 + 64 = 256,
⇒ f(0) = 0
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow3\text{x}^2+16\text{x}+16=\frac{256}{4}$
$\Rightarrow3\text{x}^2+16\text{x}-48=0$
$\Rightarrow\text{x}=-\frac{4}{3}\big(2+\sqrt{13}\big),\frac{4}{3}\big(\sqrt{13}-2\big)$
Thus, $\text{c}=\frac{-8+4\sqrt{13}}{3}\in(0,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
Hence, Lagrange's mean value theorem is verified.
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$\text{x},\text{y}\geq0$