Question

let g(x) be twice differentiable function. which of the following statements about the function

f(x)=g(x)+x^{2}

is true?

if g(x)<-2 on 0,1, then f(x) is increasing on 0,1.

if g(x)>0 on 1,10, then f(10) is the absolute maximum on 1,10.

if g(0)=0 and g(0)>0, then x=0 is a local maximum of f(x).

if g(x)<0 on -5,-1, then f(x) is increasing on -5,-1.

if g(x)<-3 on (0,2), then f(0) is the absolute minimum on 0,2.

Explanation:

Step1: Find the derivative of \(f(x)\)

Given \(f(x)=g(x)+x^{2}\), by the sum - rule of differentiation \(f^{\prime}(x)=g^{\prime}(x)+2x\).

Step2: Analyze each option

• Option 1: If \(g^{\prime}(x)< - 2\) on \([0,1]\), then \(f^{\prime}(x)=g^{\prime}(x)+2x\). For \(x\in[0,1]\), \(2x\in[0,2]\). So \(f^{\prime}(x)=g^{\prime}(x)+2x< - 2 + 2=0\) on \([0,1]\). So \(f(x)\) is decreasing on \([0,1]\), this option is false.
• Option 2: If \(g^{\prime}(x)>0\) on \([1,10]\), then \(f^{\prime}(x)=g^{\prime}(x)+2x>0\) (since \(x\in[1,10]\) and \(2x\geq2\)). So \(f(x)\) is increasing on \([1,10]\). Then \(f(10)\) is the absolute maximum on \([1,10]\) (because if a function \(y = f(x)\) is increasing on \([a,b]\), then \(f(b)=\max_{x\in[a,b]}f(x)\)).
• Option 3: If \(g^{\prime}(0) = 0\) and \(g^{\prime\prime}(0)>0\), then \(f^{\prime}(x)=g^{\prime}(x)+2x\), \(f^{\prime}(0)=g^{\prime}(0)+0 = 0\), and \(f^{\prime\prime}(x)=g^{\prime\prime}(x)+2\), \(f^{\prime\prime}(0)=g^{\prime\prime}(0)+2>0\). By the second - derivative test, \(x = 0\) is a local minimum of \(f(x)\) (since \(f^{\prime}(0) = 0\) and \(f^{\prime\prime}(0)>0\)), this option is false.
• Option 4: If \(g^{\prime}(x)<0\) on \([-5,-1]\), then \(f^{\prime}(x)=g^{\prime}(x)+2x\). For \(x\in[-5,-1]\), \(2x\in[-10,-2]\). So \(f^{\prime}(x)=g^{\prime}(x)+2x<0\) (sum of two negative numbers), \(f(x)\) is decreasing on \([-5,-1]\), this option is false.
• Option 5: If \(g^{\prime}(x)<-3\) on \((0,2)\), then \(f^{\prime}(x)=g^{\prime}(x)+2x\). For \(x\in(0,2)\), \(2x\in(0,4)\). \(f^{\prime}(x)\) can be positive or negative. For example, when \(x = 2\), \(f^{\prime}(2)=g^{\prime}(2)+4\). If \(g^{\prime}(2)=-3.5\), then \(f^{\prime}(2)=-3.5 + 4=0.5>0\). So we cannot say \(f(0)\) is the absolute minimum on \([0,2]\), this option is false.

Answer:

If \(g^{\prime}(x)>0\) on \([1,10]\), then \(f(10)\) is the absolute maximum on \([1,10]\).