Question

Use limits to give mathematical definitions for:

(a) the slope of the line tangent to the graph of a function f at the point x = 4.

(b) the line tangent to the graph of a function f at the point x = 4.

(c) the instantaneous rate of change of a function f at the point x = 1.

(d) the acceleration at time t = 1.65 of an object that moves with position function s(t).

Short answer

Part (a).$f^{\mathrm{\prime }}\left(4\right)=\underset{h\to 0}{lim} \frac{f(4+h)-f\left(4\right)}{h}$

Part (b).$f\left(x\right)=f\left(c\right)+f^{\mathrm{\prime }}\left(c\right)(x-c)$

Part (c).$f^{\mathrm{\prime }}\left(1\right)=\underset{h\to 0}{lim} \frac{f(1+h)-f\left(1\right)}{h}$

Part (d).$s^{\prime \prime }\left(1.65\right)=\underset{h\to 0}{lim} \frac{s^{\mathrm{\prime }}(1.65+h)-s^{\mathrm{\prime }}\left(1.65\right)}{h}$

Step-by-step solution

Part (a) Step 1. Given information.

We have to give mathematical definitions for the slope of the line tangent to the graph of a function f at the point x = 4 using limits.

Part (a) Step 2. Use limits to give definition

We have to find the slope of the tangent to the graph of the function f at a point x=4,

Use the principal of derivative to find the slope of the line at x=4 as shown below:

$f^{\mathrm{\prime }}\left(4\right)=\underset{h\to 0}{lim} \frac{f(4+h)-f\left(4\right)}{h}$

Part (b) Step 1. Use limits to give definition 

Use the principle of linearization to find the equation of the tangent line as shown below :

$f\left(x\right)=f\left(c\right)+f^{\mathrm{\prime }}\left(c\right)(x-c)$

Part (c) Step 1. Use limits to give definition 

Use the principle of derivative to find the instantaneous rate of change of a function at x=1 as shown below :

$f^{\mathrm{\prime }}\left(1\right)=\underset{h\to 0}{lim} \frac{f(1+h)-f\left(1\right)}{h}$

Part (d) Step 1. Use limits to give definition 

Use the principle of derivative to find the rate of change of a function s at the point t=1.65 as shown below :

$s^{\mathrm{\prime }}\left(t\right)=\underset{h\to 0}{lim} \frac{s(t+h)-s\left(t\right)}{h}$

Again s'(t) gives the velocity of the function s at a particular time.

Again differentiate the function

$s^{\prime \prime }\left(1.65\right)=\underset{h\to 0}{lim} \frac{s^{\mathrm{\prime }}(1.65+h)-s^{\mathrm{\prime }}\left(1.65\right)}{h}$