12.4 Derivatives  (Page 11/18)

$\underset{x\to c}{\lim }\left(f(x)+g(x)\right)$

$\underset{x\to c}{\lim }\frac{f(x)}{g(x)}$

$\underset{x\to c}{\lim }\left(f(x)\cdot g(x)\right)$

$\underset{x\to 0^{+}}{\lim }f(x),f(x)=\{\begin{array}{c}3x^2+2x+1\\ 5x+3\end{array}\begin{array}{c}x>0\\ x<0\end{array}$

$\underset{x\to 0^{-}}{\lim }f(x),f(x)=\{\begin{array}{c}3x^2+2x+1\\ 5x+3\end{array}\begin{array}{c}x>0\\ x<0\end{array}$

$\underset{x\to 3^{+}}{\lim }\left(3x-\mathrm{〚x〛}\right)$

$\underset{h\to 0}{\lim }\left(\frac{{\left(h+6\right)}^2-36}{h}\right)$

$\underset{x\to 25}{\lim }\left(\frac{x^2-625}{\sqrt{x}-5}\right)$

$\underset{x\to 1}{\lim }\left(\frac{-x^2-9x}{x}\right)$

$\begin{array}{c}\lim \\ {}^{x\to 4}\end{array}\frac{7-\sqrt{12x+1}}{x-4}$

$\underset{x\to -3}{\lim }\left(\frac{\frac{1}{3}+\frac{1}{x}}{3+x}\right)$

$f(x)=\frac{-2}{x-4};a=4$

$f(x)=\frac{-2}{{\left(x-4\right)}^2};a=4$

$f(x)=\frac{-x}{x^2-x-6};a=3$

$f(x)=\frac{6x^2+23x+20}{4x^2-25};a=-\frac{5}{2}$

$f(x)=\frac{\sqrt{x}-3}{9-x};a=9$

$f(x)=x^2-2x-15$

$f(x)=\frac{x^2-2x-15}{x-5}$

$f(x)=\frac{x^2-2x}{x^2-4x+4}$

$f(x)=\frac{x^3-125}{2x^2-12x+10}$

$f(x)=\frac{x^2-\frac{1}{x}}{2-x}$

$f(x)=\frac{x+2}{x^2-3x-10}$

$f(x)=\frac{x+2}{x^3+8}$

$f(x)=3x+2$

$f(x)=5$

$f(x)=\frac{1}{x+1}$

$f(x)=\ln (x)$

$f(x)=e^{2x}$

$f(x)=4x-6$

$f(x)=5x^2-3x$

Find the equation of the tangent line to the graph of $f\left(x\right)$ at the indicated $x$ value.
$f(x)=-x^3+4x$ ; $x=2.$

$f(x)=\frac{x}{\left|x\right|}$

Given that the volume of a right circular cone is $V=\frac{1}{3}\pi r^{2}h$ and that a given cone has a fixed height of 9 cm and variable radius length, find the instantaneous rate of change of volume with respect to radius length when the radius is 2 cm. Give an exact answer in terms of $\pi$

$f(1)$

$\underset{x\to {\mathrm{-1}}^{+}}{\lim }f(x)$

$\underset{x\to {\mathrm{-1}}^{-}}{\lim }f(x)$

$\underset{x\to \mathrm{-1}}{\lim }f(x)$

$\underset{x\to \mathrm{-2}}{\lim }f(x)$

At what values of $x$ is $f$ discontinuous? What property of continuity is violated?

$f(x)=\{\begin{array}{ll}\frac{1}{x}-3,\text{ i}f\hfill & x\le 2\hfill \\ x^3+1,if\hfill & x>2\hfill \end{array}a=2$

$\underset{x\to \mathrm{-5}}{\lim }\left(\frac{\frac{1}{5}+\frac{1}{x}}{10+2x}\right)$

$\underset{h\to 0}{\lim }\left(\frac{\sqrt{h^2+25}-5}{h^2}\right)$

$\underset{h\to 0}{\lim }\left(\frac{1}{h}-\frac{1}{h^2+h}\right)$

$f(x)=\sqrt{x^2-4}$

$f(x)=\frac{x^3-4x^2-9x+36}{x^3-3x^2+2x-6}$

$f(x)=\frac{3}{5+2x}$

$f(x)=\frac{3}{\sqrt{x}}$

$f(x)=2x^2+9x$

For the graph in [link] , determine where the function is continuous/discontinuous and differentiable/not differentiable.

$f(x)=\left|x-2\right|-\left|x+2\right|$

$f(x)=\frac{2}{1+e^{\frac{2}{x}}}$

$s(0)$

$s(2)$

$s\text{'}(2)$

$\frac{s(2)-s(1)}{2-1}$

$s(t)=0$

$\underset{x\to 0}{\lim }\frac{\sin (x)}{3x}$

$\underset{x\to 0}{\lim }\frac{{\tan }^2(x)}{2x}$

$\underset{x\to 0}{\lim }\frac{\sin (x)(1-\cos (x))}{2x^2}$

Evaluate the limit by hand.

$\begin{array}{c}\lim \\ {}^{x\to 1}\end{array}f(x),\text{ where }f(x)=\{\begin{array}{cc}4x-7& x\ne 1\\ x^2-4& x=1\end{array}$

At what value(s) of $x$ is the function below discontinuous?

$f(x)=\{\begin{array}{c}4x-7x\ne 1\\ x^2-4x=1\end{array}$

Find the average rate of change of the function from $x=1\text{ to }x=3.$

Find all values of $x$ at which $f\text{'}(x)=0.$

Find all values of $x$ at which $f\text{'}(x)$ does not exist.

Find an equation of the tangent line to the graph of $f$ the indicated point: $f(x)=3x^2-2x-6,x=-2$

Graph the function $f(x)=x{\left(1-x\right)}^{\frac{2}{5}}$ by entering $f(x)=x{\left({\left(1-x\right)}^2\right)}^{\frac{1}{5}}$ and then by entering $f(x)=x{\left({\left(1-x\right)}^{\frac{1}{5}}\right)}^2$ .

Explore the behavior of the graph of $f(x)$ around $x=1$ by graphing the function on the following domains, [0.9, 1.1], [0.99, 1.01], [0.999, 1.001], and [0.9999, 1.0001]. Use this information to determine whether the function appears to be differentiable at $x=1.$

$f(x)=2x-8$

$f(x)=4x^2-7$

$f(x)=x-\frac{1}{2}{x}^2$

$f(x)=\frac{1}{x+2}$

$f(x)=\frac{3}{x-1}$

$f(x)=-x^3+1$

$f(x)=x^2+x^3$

$f(x)=\sqrt{x-1}$