Question

Find the local maximum of \[ f(x)=x^3-3x+2 \]

• A. \(x=1\)
• B. \(x=-1\)
• C. \(x=0\)
• D. \(x=2\)

Hint:

If \[ f''(x)<0 \] then the function has a local maximum at that point.

Answer 1

The Correct Option is B

Concept: A local maximum or minimum occurs at points where: \[ f'(x)=0 \] These points are called critical points. To determine whether the point is maximum or minimum, we use the second derivative test.

Step 1: Find first derivative Given: \[ f(x)=x^3-3x+2 \] Differentiate with respect to \(x\): \[ f'(x)=3x^2-3 \]

Step 2: Find critical points Set: \[ f'(x)=0 \] \[ 3x^2-3=0 \] Divide by \(3\): \[ x^2-1=0 \] \[ x^2=1 \] \[ x=\pm1 \] Thus the critical points are: \[ x=1,\quad x=-1 \]

Step 3: Find second derivative Differentiate again: \[ f''(x)=6x \]

Step 4: Apply second derivative test At: \[ x=-1 \] \[ f''(-1)=6(-1)=-6 \] Since: \[ f''(-1)<0 \] the curve is concave downward. Hence: \[ x=-1 \] gives a local maximum. Final Answer: \[ \boxed{x=-1} \]