Question

find an equation of the tangent line to the curve y = x² cos x at the point (π, -π²).
o y = - 2πx
o y = 2πx
o y = -πx
o y = 2πx - 3π²
o y = - 2πx+π²

question 4
2 pts
for what value(s) of x in the interval 0, π does the graph of f have a horizontal tangent?
f(x)=2x - tan x
o x = π/3 and x = 2π/3
o x = π/6
o x = π and x = 0
o x = π/2
o x = π/4 and x = 3π/4

Explanation:

Step1: Find the derivative of $y = x^{2}\cos x$

Use the product - rule $(uv)^\prime=u^\prime v + uv^\prime$, where $u = x^{2}$ and $v=\cos x$. $u^\prime = 2x$ and $v^\prime=-\sin x$. So $y^\prime=2x\cos x - x^{2}\sin x$.

Step2: Evaluate the derivative at $x = \pi$

Substitute $x=\pi$ into $y^\prime$. $y^\prime|_{x = \pi}=2\pi\cos\pi-\pi^{2}\sin\pi$. Since $\cos\pi=- 1$ and $\sin\pi = 0$, we have $y^\prime|_{x = \pi}=2\pi\times(-1)-\pi^{2}\times0=-2\pi$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(\pi,-\pi^{2})$ and $m=-2\pi$. So $y+\pi^{2}=-2\pi(x - \pi)$. Expand to get $y=-2\pi x+\pi^{2}$.

for second part:

Step1: Find the derivative of $f(x)=2x-\tan x$

The derivative of $2x$ is $2$ and the derivative of $\tan x$ is $\sec^{2}x$. So $f^\prime(x)=2-\sec^{2}x$.

Step2: Set the derivative equal to zero

Set $f^\prime(x) = 0$, so $2-\sec^{2}x=0$. Then $\sec^{2}x = 2$, and $\sec x=\pm\sqrt{2}$. Since $\sec x=\frac{1}{\cos x}$, we have $\cos x=\pm\frac{\sqrt{2}}{2}$.

Step3: Find $x$ in the interval $[0,\pi]$

In the interval $[0,\pi]$, when $\cos x=\frac{\sqrt{2}}{2}$, $x = \frac{\pi}{4}$; when $\cos x=-\frac{\sqrt{2}}{2}$, $x=\frac{3\pi}{4}$.

Answer:

$y=-2\pi x+\pi^{2}$