Question

the function f(x) = x³+6x² + (9+2k)x +1 is strictly increasing for all x if

a) K >32
b) K <32
$\sf{c)k \geqslant 32}$
$\sf{d) \: k \leqslant 32}$
​

Answer 1

Appropriate Question:

The function $\sf f(x) = x^3+6x^2 + (9+2k)x +1$ is strictly increasing for all x if:

a) K > 3/2

b) K < 3/2

c) K ≥ 3/2

d) K ≤ 3/2

Step-by-step explanation:

Function we are provided with is:

$\sf f(x) = x^3 + 6x^2 + (9+2k)x + 1$

We have to find the value of k for which the function is strictly increasing.

Before solving this problem, we must know the concept of increasing and decreasing functions.

• If derivative of the function is greater than or equal to 0, this implies that the function is of increasing nature for all the values in its domain and vise versa in the case of decreasing function.

By differentiating the given function, we get:

${ \sf \implies f'(x) = 3x^2 + 12x + 9 + 2k}$

Now this derivative of the function must be greater than 0 for the function to be strictly increasing.

Therefore:

$\sf \implies f'(x) \geqslant 0$

${ \sf \implies 3x^2 + 12x + 9 + 2k \geqslant 0}$

${ \sf \implies 3(x^2 + 4x + 3) + 2k \geqslant 0}$

${ \sf \implies 3(x + 1)(x + 3) + 2k \geqslant 0}$

${ \sf \implies 2k \geqslant -3(x + 1)(x + 3) }$

${ \sf \implies k \geqslant \dfrac{ -3(x + 1)(x + 3)}{2} }$

Now, consider another function:

${\longrightarrow \sf g(x) = (x+1)(x+3)}$

${\longrightarrow \sf g(x) = x^2 + 4x + 3}$

Setting the derivative of the function equal to 0 will give us the minimum value because polynomial functions of odd degree are always continuous function and are increasing.

${\longrightarrow \sf g'(x) = 2x + 4}$

$\longrightarrow \sf 0 = 2x + 4$

$\longrightarrow \sf -4= 2x$

$\longrightarrow \sf -2 = x$

Therefore the minimum value of the function g(x) will be at -2. Let's find out the minimum value by plugging in x = -2.

$\longrightarrow \sf g(-2) = (-2)^2 + 4(-2) + 3$

$\longrightarrow \sf g(-2) = 4 -8 + 3$

$\boxed{ \longrightarrow \sf g(-2) =-1}$

Therefore, the value of k will be:

${ \sf \implies k \geqslant \dfrac{ -3(x + 1)(x + 3)}{2} }$

${ \sf \implies k \geqslant \dfrac{ -3( - 1)}{2} }$

${ \sf \implies k \geqslant \dfrac{3}{2} }$

Hence option (C) is correct.

Learn more:

Find the value of 'a' for which f(x)= √3 sin x−cos x−2ax+b decrease for all real values of x.

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