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