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) the derivative of a function $f$ at $x$, denoted by $f(x)$, is $f(x)=limlimits_{h\to0}\frac{f(x + h)-f(x)}{h}$ provided that the limit exists. use the\nfinition of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$.\n$f(1)=limlimits_{h\to0}\frac{(1 + h)^{2}-4-(1)^{2}-4}{h}$\nevaluate the limit expression to find $f(1)$.\n$f(1)=2$ (type an integer or a fraction.)\n(b) $f(1)=-3$ (type an integer or a fraction)\nthe equation of the tangent line at $(1, f(1))$ is $2x - 5$. (type your answer in slope - intercept form.)

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) the derivative of a function $f$ at $x$, denoted by $f(x)$, is $f(x)=limlimits_{h\to0}\frac{f(x + h)-f(x)}{h}$ provided that the limit exists. use the\nfinition of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$.\n$f(1)=limlimits_{h\to0}\frac{(1 + h)^{2}-4-(1)^{2}-4}{h}$\nevaluate the limit expression to find $f(1)$.\n$f(1)=2$ (type an integer or a fraction.)\n(b) $f(1)=-3$ (type an integer or a fraction)\nthe equation of the tangent line at $(1, f(1))$ is $2x - 5$. (type your answer in slope - intercept form.)

Answer

Explanation:

Step1: Find (f(1))

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

Step2: Find the derivative (f^{\prime}(1))

[ \begin{align*} f^{\prime}(1)&=\lim_{h\rightarrow0}\frac{f(1 + h)-f(1)}{h}\ &=\lim_{h\rightarrow0}\frac{(1 + h)^{2}-4-(1^{2}-4)}{h}\ &=\lim_{h\rightarrow0}\frac{1 + 2h+h^{2}-4 - 1 + 4}{h}\ &=\lim_{h\rightarrow0}\frac{2h+h^{2}}{h}\ &=\lim_{h\rightarrow0}(2 + h)\ &=2 \end{align*} ]

Step3: Find the equation of the tangent line

The slope - intercept form of a line is (y=mx + b), where (m) is the slope and (b) is the (y) - intercept. We know that (m = f^{\prime}(1)=2) and the line passes through the point ((1,f(1))=(1,-3)). Substitute (x = 1), (y=-3) and (m = 2) into (y=mx + b) (-3=2\times1+b) (b=-3 - 2=-5) The equation of the tangent line is (y = 2x-5)

Answer:

(a) (f^{\prime}(1)=2) (b) (f(1)=-3), equation of the tangent line (y = 2x-5)