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  • The interval on which the function f (x) equals 2x3 plus 9x2 plus 12x minus 1 is decreasing is: A. [minus1, ∞)

The interval on which the function f (x) = 2x^3 + 9x^2 + 12x - 1 is decreasing is:
A. [-1, ∞)
B. [-2, -1]
C. (-∞, -2]
D. [-1, 1]

Answers (1)

Given f (x) = 2x^3 + 9x^2 + 12x - 1

Apply first derivative and get

\begin{aligned} &f(x)=\frac{d\left(2 x^{3}+9 x^{2}+12 x-1\right)}{d x}\\ &\text { Apply the sum rule of the differentiation and } 0 \text { is the derivative of the constant, so }\\ &\Rightarrow f(x)=\frac{d\left(2 x^{3}\right)}{d x}+\frac{d\left(9 x^{2}\right)}{d x}+\frac{d(12 x)}{d x}-\frac{d(1)}{d x} \end{aligned}

Apply power rule and get

\\ { \Rightarrow f'(x)=6x\textsuperscript{2}+18x+12-0}\\ { \Rightarrow f'(x)=6(x\textsuperscript{2}+3x+2)}\\

Split the middle term and get

\\ { \Rightarrow f'(x)=6(x\textsuperscript{2}+2x+x+2)}\\ { \Rightarrow f'(x)=6(x(x+2)+1(x+2))}\\ { \Rightarrow f'(x)=6((x+2) (x+1))}\\

Now f'(x)=0 gives

x=-1, -2

Three intervals are made when these points divide the real number line

\\ {(- \infty , -2), [-2,-1] and (-1, \infty )}\\ ${(i) in the interval $(- \infty , -2), f'(x)>0}\\ { \therefore $ f(x) is increasing in (- \infty ,-2)}\\ ${(ii) $in the interval [-2,-1], f'(x) \leq 0}\\ ${ \therefore $ f(x) is decreasing in [-2, -1]}$ \\ $(iii) in the interval$ (-1, \infty ), f'(x)>0\\ { \therefore $ f(x) is increasing in (-1, \infty )}\\

So, the interval on which the function decreases is [-2, -1].

So, the correct answer is option B.

Posted by

infoexpert22

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