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Find the Intervals in Which the Function F ( X ) = 3 2 X 4 − 4 X 3 − 45 X 2 + 51 is (A) Strictly Increasing (B) Strictly Decreasing

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Question

Find the intervals in which the function \[f(x) = \frac{3}{2} x^4 - 4 x^3 - 45 x^2 + 51\] is

(a) strictly increasing
(b) strictly decreasing

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Solution 1

Given:\[f(x) = \frac{3}{2} x^4 - 4 x^3 - 45 x^2 + 51\]

Differentiating w.r.t. x, we get:
f'(x) = \[6 x^3 - 12 x^2 - 90x\]

\[6x\left( x^2 - 2x - 15 \right)\] At critical points, f'(x)=0.

\[6x\left( x^2 - 2x - 15 \right)\] =0

\[\Rightarrow 6x\left( x^2 - 5x + 3x - 15 \right) = 0\]

\[ \Rightarrow 6x\left( x - 5 \right)\left( x + 3 \right) = 0\]

\[ \Rightarrow x = - 3, 0, 5\]

 

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Solution 2

Given:\[f(x) = \frac{3}{2} x^4 - 4 x^3 - 45 x^2 + 51\]

Differentiating w.r.t. x, we get:
f'(x) = \[6 x^3 - 12 x^2 - 90x\]

\[6x\left( x^2 - 2x - 15 \right)\] At critical points, f'(x)=0.

\[6x\left( x^2 - 2x - 15 \right)\] =0

\[\Rightarrow 6x\left( x^2 - 5x + 3x - 15 \right) = 0\]

\[ \Rightarrow 6x\left( x - 5 \right)\left( x + 3 \right) = 0\]

\[ \Rightarrow x = - 3, 0, 5\]

Interval f'(x)= \[6x\left( x - 5 \right)\left( x + 3 \right)\] Result
\[\left( - \infty , - 3 \right)\] f'(-4)=-216 <0 strictly decreasing
\[\left( - 3, 0 \right)\] f'(-1)=  72 >0 strictly increasing
\[\left( 0, 5 \right)\] f'(1)= -96 <0 strictly decreasing
\[\left( 5, \infty \right)\] f'(6)=324 >0 strictly increasing
 

(a) Hence the function is strictly increasing in \[\left( - 3, 0 \right)\] \[\cup\] \[\left( 5, \infty \right)\] .

(b) Also, the function is strictly decreasing in \[\left( - \infty , - 3 \right)\] \[\cup\] \[\left( 0, 5 \right)\] .

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Solution 3

Given:\[f(x) = \frac{3}{2} x^4 - 4 x^3 - 45 x^2 + 51\]

Differentiating w.r.t. x, we get:
f'(x) = \[6 x^3 - 12 x^2 - 90x\]

\[6x\left( x^2 - 2x - 15 \right)\] At critical points, f'(x)=0.

\[6x\left( x^2 - 2x - 15 \right)\] =0

\[\Rightarrow 6x\left( x^2 - 5x + 3x - 15 \right) = 0\]

\[ \Rightarrow 6x\left( x - 5 \right)\left( x + 3 \right) = 0\]

\[ \Rightarrow x = - 3, 0, 5\]

Interval f'(x)= \[6x\left( x - 5 \right)\left( x + 3 \right)\] Result
\[\left( - \infty , - 3 \right)\] f'(-4)=-216 <0 strictly decreasing
\[\left( - 3, 0 \right)\] f'(-1)=  72 >0 strictly increasing
\[\left( 0, 5 \right)\] f'(1)= -96 <0 strictly decreasing
\[\left( 5, \infty \right)\] f'(6)=324 >0 strictly increasing
 

(a) Hence the function is strictly increasing in \[\left( - 3, 0 \right)\] \[\cup\] \[\left( 5, \infty \right)\] .

(b) Also, the function is strictly decreasing in \[\left( - \infty , - 3 \right)\] \[\cup\] \[\left( 0, 5 \right)\] .

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2013-2014 (March) Foreign Set 1

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