use the function below to answer parts (a)-(c).\n$f(x)=x^{2}-4$\n(a) use the formal definition to find the…

use the function below to answer parts (a)-(c).\n$f(x)=x^{2}-4$\n(a) use the formal definition to find the derivative of $y = f(x)$ at $x = 1$.\n(b) find $f(1)$ and find the equation of the tangent line at the point $(1, f(1))$.\n(c) graph $y = f(x)$ and the tangent line at the point $(1, f(1))$ in the same coordinate system.\n(a) $f1 = - 3$ (type an integer or a fraction.)\n(b) $f(1) = square$, and the equation of the tangent line at $(1, f(1))$ is $y = 2x - 5$. (type your answer in slope-intercept form.)\n(c) choose the correct graph below.\na.\nb.\nc.\nd.

use the function below to answer parts (a)-(c).\n$f(x)=x^{2}-4$\n(a) use the formal definition to find the derivative of $y = f(x)$ at $x = 1$.\n(b) find $f(1)$ and find the equation of the tangent line at the point $(1, f(1))$.\n(c) graph $y = f(x)$ and the tangent line at the point $(1, f(1))$ in the same coordinate system.\n(a) $f1 = - 3$ (type an integer or a fraction.)\n(b) $f(1) = square$, and the equation of the tangent line at $(1, f(1))$ is $y = 2x - 5$. (type your answer in slope-intercept form.)\n(c) choose the correct graph below.\na.\nb.\nc.\nd.

Answer

Explanation:

Step1: Find the derivative using the formal definition

The formal definition of the derivative is (f^\prime(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}). Given (f(x)=x^{2}-4), then (f(x + h)=(x + h)^{2}-4=x^{2}+2xh+h^{2}-4). [ \begin{align*} f^\prime(x)&=\lim_{h\rightarrow0}\frac{(x^{2}+2xh + h^{2}-4)-(x^{2}-4)}{h}\ &=\lim_{h\rightarrow0}\frac{x^{2}+2xh+h^{2}-4 - x^{2}+4}{h}\ &=\lim_{h\rightarrow0}\frac{2xh+h^{2}}{h}\ &=\lim_{h\rightarrow0}(2x + h) \end{align*} ] As (h\rightarrow0), (f^\prime(x)=2x). At (x = 1), (f^\prime(1)=2\times1=2).

Step2: Find (f(1))

Substitute (x = 1) into (f(x)=x^{2}-4), so (f(1)=1^{2}-4=-3).

Step3: Find the equation of the tangent line

The point - slope form of a line is (y - y_{1}=m(x - x_{1})), where ((x_{1},y_{1})=(1,-3)) and (m = 2). [ \begin{align*} y-(-3)&=2(x - 1)\ y+3&=2x-2\ y&=2x-5 \end{align*} ]

Answer:

(a) The derivative of (y = f(x)) at (x = 1) is (2). (b) (f(1)=-3), and the equation of the tangent line is (y = 2x-5). (c) To graph (y=f(x)=x^{2}-4) (a parabola opening upwards with vertex ((0,-4))) and (y = 2x-5) (a straight line with slope (2) and (y) - intercept (-5)), plot points for the parabola (e.g., when (x = 0,y=-4); (x=2,y = 0); (x=-2,y = 0)) and for the line (e.g., when (x = 0,y=-5); when (y = 0,x=\frac{5}{2})). The correct graph is the one where the line (y = 2x - 5) touches the parabola (y=x^{2}-4) at the point ((1,-3)).