Question

find all critical number(s) of
g(x)=int_{0}^{-3 x^{3}+4 x^{2}} e^{-t^{2}} d t
if there are fewer than two critical numbers, leave the remaining fields blank.
( x=quad ) number (3 significant figures)
( x=quad ) number (3 significant figures)

Explanation:

Step1: Apply the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus, if \(g(x)=\int_{a}^{u(x)}f(t)dt\), then \(g^{\prime}(x)=f(u(x))\cdot u^{\prime}(x)\). Here \(a = 0\), \(u(x)=-3x^{3}+4x^{2}\), and \(f(t)=e^{-t^{2}}\). So \(g^{\prime}(x)=e^{-(-3x^{3}+4x^{2})^{2}}\cdot(-9x^{2}+8x)\).

Step2: Find where \(g^{\prime}(x) = 0\)

Since \(e^{-(-3x^{3}+4x^{2})^{2}}>0\) for all real \(x\) (because the exponential function \(y = e^{z}\) is positive for all \(z\in R\)), we set \(-9x^{2}+8x=0\). Factor out an \(x\): \(x(-9x + 8)=0\).

Step3: Solve the equation \(x(-9x + 8)=0\)

Using the zero - product property \(ab = 0\) implies \(a = 0\) or \(b=0\).

• If \(x=0\), then the equation is satisfied.
• If \(-9x + 8=0\), then \(9x=8\), so \(x=\frac{8}{9}\approx0.889\).

Answer:

\(x = 0\)
\(x\approx0.889\)