Question

question 9 (2 points)
the function ( f(x)=x^{3}-48 x ) has a local maximum at ( x= ), and has a local minimum at ( x= ). note that here we are looking for the ( x ) value (not the range value). enter a number for each answer.
view hint for question 9
question 10 (1 point)
suppose the second derivative of a function ( f(x) ) is ( f^{prime prime}(x)=2 x-1 ). suppose ( x=1 ) is a critical point of ( f ). using the second derivative test, we know ( x=1 ) is

• a local maximum
• a local minimum
• an absolute minimum
• an absolute maximum

view hint for question 10

Explanation:

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

The function is \(f(x)=x^{3}-48x\). Using the power rule \((x^n)^\prime = nx^{n - 1}\), the first derivative \(f^\prime(x)=3x^{2}-48\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), so \(3x^{2}-48=0\). Factor out 3: \(3(x^{2}-16)=0\), which gives \(x^{2}-16=(x + 4)(x - 4)=0\). The critical points are \(x=-4\) and \(x = 4\).

Step3: Find the second derivative of \(f(x)\)

Differentiate \(f^\prime(x)=3x^{2}-48\) with respect to \(x\). Using the power rule, \(f^{\prime\prime}(x)=6x\).

Step4: Use the second - derivative test

• For \(x=-4\): Substitute \(x=-4\) into \(f^{\prime\prime}(x)\), \(f^{\prime\prime}(-4)=6\times(-4)=-24<0\). So \(x = - 4\) is a local maximum.
• For \(x = 4\): Substitute \(x = 4\) into \(f^{\prime\prime}(x)\), \(f^{\prime\prime}(4)=6\times4 = 24>0\). So \(x = 4\) is a local minimum.

For Question 10:

Step1: Apply the second - derivative test

Given \(f^{\prime\prime}(x)=2x - 1\) and \(x = 1\) is a critical point. Substitute \(x = 1\) into \(f^{\prime\prime}(x)\), \(f^{\prime\prime}(1)=2\times1-1=1>0\).

Answer:

Question 9: local maximum at \(x=-4\), local minimum at \(x = 4\).
Question 10: a local minimum.