Question

find all values of x where the function ( f(x)=\frac{-2x}{e^{2x^{2}-3x}} ) has a relative extrema.

Explanation:

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

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Let \(u = - 2x\), \(u^\prime=-2\) and \(v = e^{2x^{2}-3x}\), \(v^\prime=(4x - 3)e^{2x^{2}-3x}\).

Then \(f^\prime(x)=\frac{-2e^{2x^{2}-3x}-(-2x)(4x - 3)e^{2x^{2}-3x}}{(e^{2x^{2}-3x})^{2}}=\frac{e^{2x^{2}-3x}(-2 + 8x^{2}-6x)}{e^{4x^{2}-6x}}=\frac{8x^{2}-6x - 2}{e^{2x^{2}-3x}}\)

Step2: Set \(f^\prime(x)=0\)

Set the numerator equal to zero: \(8x^{2}-6x - 2 = 0\). Divide through by \(2\): \(4x^{2}-3x - 1=0\)

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here \(a = 4\), \(b=-3\), \(c=-1\)

\(x=\frac{3\pm\sqrt{(-3)^{2}-4\times4\times(-1)}}{2\times4}=\frac{3\pm\sqrt{9 + 16}}{8}=\frac{3\pm5}{8}\)

Step3: Solve for \(x\)

For \(x=\frac{3 + 5}{8}\), \(x = 1\)

For \(x=\frac{3-5}{8}\), \(x=-\frac{1}{4}\)

Answer:

The function \(f(x)\) has relative extrema at \(x = 1\) and \(x=-\frac{1}{4}\)