Question

Suppose that f(x)=3x2−4.


(A) Find the slope of the line tangent to f(x) at x=−1.

(B) Find the instantaneous rate of change of f(x) at x=−1.

(C) Find the equation of the line tangent to f(x) at x=−1. y=

Answer 1

(A) Given: $f(x)=$ $3x^2-4$

Require to find the slope of the line tangent to $f(x)$ at $x=-1$

Find the derivative of the given function $f(x)$ = $3x^2 - 4$

Now $f(x)=$ $3x^2 - 4$ $\Rightarrow f'(x)=3(2x) - 0 = 6x$ (Using Power Rule)

Slope of the line tangent to $f(x)$ at $x=-1$ is given by

$m=f'(-1)=6(-1)=-6$

Therefore, slope of the line tangent to $f(x)$ at $x=-1$ is $m=-6$

(B)Require to find the instantaneous rate of change of $f(x)$ at $x=-1$

Instantaneous rate of change of $f(x)$ at $x=-1$ is given by $f'(-1)$ and

$f'(-1)=-6$

Therefore, instantaneous rate of change = - 1

(C)Require to find the equation of the line tangent to $f(x)$ at $x=-1$

From part(A), we have slope of the line tangent to $f(x)$ at $x=-1$ is $m=-6$

And $x=-1$ $\Rightarrow f(-1)=3(-1)^2- 4 = 3-4=-1$

Now let us find the equation of the line tangent to $f(x)$ at the point (- 1, - 1) using point-slope form.

Equation of the line tangent to f(x) in point-slope form is

$y-y_{1}=m(x-x_{1})$ , where $(x_{1},y_{1})= (-1, -1)$ and $m=-6$

Using the above, equation of the line tangent to f(x) at x = - 1 is

$y-(-1)=-6(x-(-1))$

That is, $y+1=-6(x+1)$

That is, $y=-6x-6-1$

Therefore, equation of the line tangent to the given function is

$y=-6x-7$