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Determine if the $x$-axis is tangent to $f(x) = x^2$ at the origin.
Any angle drawn with vertex at the origin that contains the $x$-axis has one side above the $x$-axis.
This side must be a line distinct from the $x$-axis that also passes through the origin and thus must take the form $y=mx$ for some $m \gt 0$.Then, for any positive $x_0 \lt m$, upon multiplying both sides of this inequality by $x_0$, we have $0 \lt x_0^2 \lt mx_0$.
This means that a point on the graph of $f(x)$, namely $(x_0,x_0^2)$, is between the $x$-axis (i.e., $y=0$) and a point on the line $y=mx$. In this way, we have found a point on the curve of $f(x)$ in the interior of the aforementioned angle. That is to say, the curve defined by $f(x) = x^2$ always enters such an angle, no matter the angle (of non-zero measure) considered.
Thus, the $x$-axis is indeed tangent to $f(x)$.
Determine if the $x$-axis is tangent to $\displaystyle{f(x) = \sin \left( \frac{1}{x} \right)}$
For a line to be tangent to a curve at some point $P$, the line must first intersect the curve at $P$. In the case of this function, note there is a domain issue at $x=0$. As such, $f$ can't possibly have a tangent there. No line, $x$-axis or otherwise, can function as a tangent at the origin for this function.
Let $s(t) = t^2 - 3t + 4$ denote the position of a particle in meters at time $t$ in seconds. What is the instantaneous velocity of the particle at any time $t$?
Consider two times, $t$ and $t + h$ for some small $h$. The positions of the particle at these two times are then given by $s(t) = t^2 - 3t + 4$ and $s(t+h) = (t + h)^2 - 3(t + h) + 4$.
The change in position over the change in time (i.e., the average velocity over these $h$ seconds) is then
$$v_{avg} = \frac{[(t + h)^2 - 3(t + h) + 4] - [t^2 - 3t + 4]}{(t + h) - t}$$which simplifies to
$$v_{avg} = \frac{2th + h^2 - 3h}{h}$$While the above expression is undefined when $h = 0$, it is identical to $2th - 3$ for all other values of $h$ (consider what things change and stay the same when the common factor of $h$ is cancelled from the numerator and denominator).
Thus, by considering small enough $h \ne 0$ (i.e., very small intervals of time), we see that the average velocity can be made as close as we want to $2t-3$.
So the instantaneous velocity of the particle at any time $t$ is given by $2t-3$ meters per second.
Let $s(t) = \sqrt{t}$ denote the position of a particle in meters at time $t$ in seconds. What is the instantaneous velocity of the particle at $t=4$ seconds?
Consider two times, $t=4$ and $t = 4+h$ for some small $h$. The positions of the particle at these two times are then given by $s(4) = 2$ and $s(4+h) = \sqrt{4+h}$.
The change in position over the change in time (i.e., the average velocity over these $h$ seconds) is then given by
$$v_{avg} = \frac{\sqrt{4+h} - 2}{h}$$It is less clear what value this approaches as $h$ gets close to zero, until we multiply by a well-chosen value of one involving the conjugate of the numerator:
$$\begin{array}{rcl} v_{avg} &=& \displaystyle{\frac{\sqrt{4+h} - 2}{h} \cdot \frac{\sqrt{4+h} + 2}{\sqrt{4+h} + 2}}\\ &=& \displaystyle{\frac{(4+h)-4}{h (\sqrt{4+h} + 2)}}\\ &=& \displaystyle{\frac{h}{h(\sqrt{4+h} + 2)}} \end{array}$$While the above expression is undefined when $h=0$, it is identical to $\displaystyle{\frac{1}{\sqrt{4+h} + 2}}$ for all other values of $h$ (consider what things change and stay the same when the common factor of $h$ is cancelled from the numerator and denominator).
Thus, by considering small enough $h \ne 0$ (i.e., very small intervals of time), we see that the average velocity can be made as close as we want to $\displaystyle{\frac{1}{\sqrt{4+0} + 2} = \frac{1}{4}}$.
Thus, the instantaneous velocity of the particle at $t=4$ seconds is $1/4$ meters per second.
Find the slope of the secant line of $f(x) = x^3$ through the points $(2,f(2))$ and $(2+h,f(2+h))$. Determine the value the slope of the secant line approaches as $h$ gets close to zero; use it to find the equation of the tangent line.
Slope of secant line: $m = h^2 + 6h + 12$; equation of tangent line (e.g.): $y - 8 = 12(x-2)$
Find the slope of the secant line of $\displaystyle{f(x) = \frac{x}{x-3}}$ through the points $(-3,f(-3))$ and $(-3+h,f(-3+h))$. Determine the value the slope of the secant line approaches as $h$ gets close to zero; use it to find the equation of the tangent line.
Slope of secant line: $\displaystyle{m = \frac{1}{2(h-6)}}$; equation of tangent line (e.g.): $\displaystyle{y - \frac{1}{2} = -\frac{1}{12}(x+3)}$
Find the instantaneous velocity of a particle at time $t=4$ if the position of a particle is given by $s(t) = t^3 - 3t$ meters at $t$ seconds.
$45$ m/s
Find the instantaneous velocity $v(t)$ of a particle at time $t$, if the position of the particle is given by $\displaystyle{s(t) = t + \frac{2}{2t+1}}$ meters at $t$ seconds.