Question

Let domain and range of \(f(x)\) and \(g(x)\) is \([0,\infty)\). If \(f(x)\) is an increasing function, \(g(x)\) is a decreasing function, \(h(x)=f\{g(x)\}\), \(h(0)=0\) and \(p(x)=h(x^{3}-2x^{2}+2x)-h(4)\) then for all \(x\in(0,2)\):

• A. $p(x)=-3$
• B. $p(x)=0$
• C. $0
• D. $0\le p(x)\le-h(4)$

Hint:

Remember this handy composite rule: composing two functions with the same monotonicity (both increasing or both decreasing) creates an *increasing* function, while composing two functions with different monotonicity creates a *decreasing* function!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to determine the monotonicity of the composite function \( h(x) \) and the range of the cubic argument \( t(x) = x^3 - 2x^2 + 2x \) in the interval \( (0, 2) \).
Step 2: Key Formula or Approach:
1. If \( f \) is increasing and \( g \) is decreasing, \( h = f \circ g \) is decreasing.
2. Analyze \( t(x) = x^3 - 2x^2 + 2x \) for monotonicity and bounds.
Step 3: Detailed Explanation:
Since \( f \) is increasing (\( f'>0 \)) and \( g \) is decreasing (\( g'<0 \)), the chain rule gives \( h'(x) = f'(g(x))g'(x) = (+) \cdot (-)<0 \). Thus, \( h(x) \) is a strictly decreasing function.
Given \( h(0) = 0 \). Since \( h \) decreases, for any \( x>0, h(x)<h(0) = 0 \).
Now let \( t = x^3 - 2x^2 + 2x \). Its derivative is \( t' = 3x^2 - 4x + 2 \).
The discriminant of \( t' \) is \( D = 16 - 24 = -8<0 \), meaning \( t' \) is always positive. So \( t(x) \) is strictly increasing.
For \( x \in (0, 2) \): - \( t(0) = 0 \)
- \( t(2) = 8 - 8 + 4 = 4 \)
So for \( x \in (0, 2) \), we have \( 0<t<4 \).
Since \( h \) is decreasing, applying \( h \) to the inequality \( 0<t<4 \) reverses it: \[ h(0)>h(t)>h(4) \implies 0>h(t)>h(4) \] We want to find the range of \( p(x) = h(t) - h(4) \). Subtract \( h(4) \) from the inequality: \[ 0 - h(4)>h(t) - h(4)>h(4) - h(4) \] \[ -h(4)>p(x)>0 \] Step 4: Final Answer:
For \( x \in (0, 2) \), we have \( 0<p(x)<-h(4) \).