Question

find f(3) given f(x)=e^{2x^{2}-2x}cos(πx). please simplify your answer. f(3)= symbolic expression save & grade 5 attempts left save only

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u(x)v(x)$, then $y'=u'(x)v(x)+u(x)v'(x)$. Let $u(x)=e^{2x^{2}-2x}$ and $v(x)=\cos(\pi x)$.

Step2: Differentiate $u(x)$

Using the chain - rule, if $y = e^{g(x)}$, then $y'=e^{g(x)}g'(x)$. For $u(x)=e^{2x^{2}-2x}$, $g(x)=2x^{2}-2x$, and $g'(x)=(4x - 2)$. So, $u'(x)=(4x - 2)e^{2x^{2}-2x}$.

Step3: Differentiate $v(x)$

If $v(x)=\cos(\pi x)$, using the chain - rule, if $y=\cos(h(x))$, then $y'=-\sin(h(x))h'(x)$. Here, $h(x)=\pi x$ and $h'(x)=\pi$. So, $v'(x)=-\pi\sin(\pi x)$.

Step4: Find $f'(x)$

By the product - rule, $f'(x)=u'(x)v(x)+u(x)v'(x)=(4x - 2)e^{2x^{2}-2x}\cos(\pi x)-\pi e^{2x^{2}-2x}\sin(\pi x)=e^{2x^{2}-2x}[(4x - 2)\cos(\pi x)-\pi\sin(\pi x)]$.

Step5: Evaluate $f'(3)$

Substitute $x = 3$ into $f'(x)$.
First, calculate $2x^{2}-2x$ when $x = 3$: $2\times3^{2}-2\times3=2\times9 - 6=18 - 6 = 12$.
Then, $\cos(3\pi)=-1$ and $\sin(3\pi)=0$.
$f'(3)=e^{12}[(4\times3 - 2)\times(-1)-\pi\times0]=e^{12}(10\times(-1))=- 10e^{12}$.

Answer:

$-10e^{12}$