Question:

Let $x=2$ be a local minima of the function $f(x)=2 x^4-18 x^2+8 x+12$, $x \in(-4,4)$ If $M$ is local maximum value of the function $f$ in $(-4,4)$, then $M =$

Updated On: Mar 20, 2025

$18 \sqrt{6}-\frac{33}{2}$
$12 \sqrt{6}-\frac{33}{2}$
$12 \sqrt{6}-\frac{31}{2}$
$18 \sqrt{6}-\frac{31}{2}$

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The Correct Option is B

Solution and Explanation

f^{'} (x) = 8 x^{3} - 36 x + 8 = 4 (2 x^{3} - 9 x + 2)

f^{'} (x) = 0

∴ x = \frac{6 - 2}{2}

Now

f (x) = (x^{2} - 2 x - \frac{9}{2}) (2 x^{2} + 4 x - 1) + 24 x + 7.5

∴ f (\frac{6 - 2}{2}) = M = 126 - \frac{33}{2}

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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by

$\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}$

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

If for any two points x₁ and x₂ in the interval x such a manner that x₁ < x₂, there holds an inequality f(x₁) ≤ f(x₂); then the function f(x) is known as increasing in this interval.
Likewise, if for any two points x₁ and x₂ in the interval x such a manner that x₁ < x₂, there holds an inequality f(x₁) ≥ f(x₂); then the function f(x) is known as decreasing in this interval.
The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x₁) < f(x₂) for strictly increasing and f(x₁) > f(x₂) for strictly decreasing.

Read More: Application of Derivatives