2 quick questions:
Find the equation of the line tangent to the curve y^2+xy+x^3=7 at the point x=2.
Given the function f(x)=ln(3+sinx) on the interval [0,7]
a. at which values of x does f have local max and local min
do you know how to differentiate implicitly? you can do that and solve for y'. you can use that to find the slope. then the equation of the tangent line is given by $\displaystyle y - y_1 = m(x - x_1)$, just solve for y. Here, $\displaystyle m$ is the slope, given by the derivative that you found earlier, $\displaystyle (x_1,y_1)$ is a point the line passes through. it will be $\displaystyle (2,y_1)$ you can solve for $\displaystyle y_1$ using the initial equation, you'll need to do this anyway to get the slope.
find $\displaystyle f'(x)$ and $\displaystyle f''(x)$.Given the function f(x)=ln(3+sinx) on the interval [0,7]
a. at which values of x does f have local max and local min
To find the critical points, set $\displaystyle f'(x) = 0$ and solve for $\displaystyle x$ (only take the values between 0 and 7 inclusive.
To find whether the point you found above are max's or min's, check the following:
If $\displaystyle f''(a) < 0$ then we have a local max
If $\displaystyle f''(a) > 0$ then we have a local min
when $\displaystyle x = 2$, we have:
$\displaystyle y^2 + 2y + 8 = 7$
$\displaystyle \Rightarrow y^2 + 2y + 1 = 0$
$\displaystyle \Rightarrow (y + 1)^2 = 0$
$\displaystyle \Rightarrow y = -1$
So that we are concerned with the point $\displaystyle (2,-1)$, this will be our $\displaystyle (x_1,y_1)$.
Now, $\displaystyle y^2 + xy + x^3 = 7$
$\displaystyle \Rightarrow 2y~y' + y + x~y' + 3x = 0$
$\displaystyle \Rightarrow (2y + x)y' = -(3x + y)$
$\displaystyle \Rightarrow y' = - \frac {3x + y}{2y + x}$
at $\displaystyle (2,-1)$, the slope is undefined. So that the tangent line is just $\displaystyle x = 2$
$\displaystyle f(x) = \ln (3 + \sin x)$Given the function f(x)=ln(3+sinx) on the interval [0,7]
a. at which values of x does f have local max and local min
$\displaystyle \Rightarrow f'(x) = \frac {\cos x}{3 + \sin x}$ ............by the chain rule
$\displaystyle \Rightarrow f''(x) = \frac {-3 \sin x - \sin^2 x - \cos^2 x }{(3 + \sin x)^2} = \frac {-3 \sin x - 1}{(3 + \sin x)^2}$
$\displaystyle f'(x) = 0 \implies \cos x = 0 \implies x = \frac {\pi}2,~\frac {3 \pi}2$ for $\displaystyle 0 \le x \le 7$
$\displaystyle f''\left( \frac {\pi}2 \right) < 0 $, so we have a local max at $\displaystyle x = \frac {\pi}2$
$\displaystyle f'' \left( \frac {3 \pi}2 \right) > 0$, so we have a local min at $\displaystyle x = \frac {3 \pi}2$