find f(x) and find the equation of the line tangent to the graph of f at the indicated value of x. find the…

Explanation:

Step1: Apply the sum - difference rule and constant - multiple rule

The sum - difference rule states that $(u\pm v\pm w)' = u'\pm v'\pm w'$ and the constant - multiple rule states that $(cf(x))'=cf'(x)$. Let $u = 2e^{x^{2}}$, $v = 5x$, and $w = 8$. Then $f(x)=u - v+w$, and $f'(x)=(2e^{x^{2}})'-(5x)'+(8)'$.

Step2: Differentiate each term

For $(2e^{x^{2}})$: Use the chain - rule. If $y = 2e^{u}$ and $u=x^{2}$, then $\frac{dy}{du}=2e^{u}$ and $\frac{du}{dx}=2x$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=2e^{u}\cdot2x = 4xe^{x^{2}}$. The derivative of $5x$ is $5$ (since $(ax)' = a$ for a constant $a$), and the derivative of a constant $8$ is $0$ (since $(c)' = 0$ for a constant $c$). So $f'(x)=4xe^{x^{2}}-5$.

Step3: Find the slope of the tangent line at $x = 0$

Substitute $x = 0$ into $f'(x)$. $f'(0)=4\times0\times e^{0^{2}}-5=-5$.

Step4: Find the value of $f(0)$

Substitute $x = 0$ into $f(x)$: $f(0)=2e^{0^{2}}-5\times0 + 8=2\times1+8=10$.

Step5: Find the equation of the tangent line

Use the point - slope form of a line $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(0,10)$ and $m=-5$. So $y - 10=-5(x - 0)$, which simplifies to $y=-5x + 10$.

Step6: Find where the tangent line is horizontal

Set $f'(x)=0$. So $4xe^{x^{2}}-5 = 0$, or $4xe^{x^{2}}=5$. Since $e^{x^{2}}>0$ for all real $x$, when $x = 0$, $4xe^{x^{2}}=0\neq5$. So there are no real values of $x$ for which the tangent line is horizontal.

Answer:

$f'(x)=4xe^{x^{2}}-5$; The equation of the tangent line at $x = 0$ is $y=-5x + 10$; There are no real values of $x$ for which the tangent line is horizontal.