Chapter 14, Problem 14CT

(a)

To determine

To calculate: The derivative of function $f(x)=4x^2-11x-3$ at $x=2$.

Answer to Problem 14CT

Solution:

The derivative of the function at $x=2$ is $f\text{'}(2)=5$.

Explanation of Solution

Given information:

The function, $f(x)=4x^2-11x-3$.

Formula used:

The derivative of the function $f(x)$ at $x=c$ is given by,

$f\text{'}(c)=\underset{x\to c}{\lim }\frac{f(x)-f(c)}{x-c}$.

Explanation:

As the function is $f(x)=4x^2-11x-3$ and $x=2$,

Therefore, $f(2)=4{(2)}^2-11(2)-3$.

$\Rightarrow f(2)=16-22-3=-9$.

By using definition of derivative,

$\Rightarrow f\text{'}(2)=\underset{x\to 2}{\lim }\frac{f(x)-f(2)}{x-2}$

$\Rightarrow f\text{'}(2)=\underset{x\to 2}{\lim }\frac{4x^2-11x-3-(-9)}{x-2}$

$\Rightarrow f\text{'}(2)=\underset{x\to 2}{\lim }\frac{4x^2-11x-3+9}{x-2}$

$\Rightarrow f\text{'}(2)=\underset{x\to 2}{\lim }\frac{4x^2-11x+6}{x-2}$.

Simplifying numerator,

$\Rightarrow f\text{'}(2)=\underset{x\to 2}{\lim }\frac{(4x-3)(x-2)}{x-2}$

$\Rightarrow f\text{'}(2)=\underset{x\to 2}{\lim }(4x-3)$.

$\Rightarrow f\text{'}(2)=5$.

(b)

To determine

An equation of tangent line to the graph of $f(x)=4x^2-11x-3$ at the point $(2,-9)$.

Answer to Problem 14CT

Solution:

The equation of tangent line to the graph of $f(x)=4x^2-11x-3$ at the point $(2,-9)$ is $y=5x-19$.

Explanation of Solution

Given information:

The function, $f(x)=4x^2-11x-3$.

Explanation:

To find an equation of tangent line at point $(2,-9)$,

The slope of the tangent line to the graph of $f(x)=4x^2-11x-3$ at $(2,-9)$ is $m_{\text{tan}}=f\text{'}(2)$

$\Rightarrow m_{\text{tan}}=5$

Hence the slope of the tangent line is $5$.

The equation of tangent line, $y-f(2)=m_{\text{tan}}(x-2)$

$\Rightarrow y-(-9)=5(x-2)$

$\Rightarrow y+9=5x-10$

Subtracting $9$ from both sides of an equations,

$\Rightarrow y=5x-19$

Hence, the equation of tangent line to the graph of $f(x)=4x^2-11x-3$ at the point $(2,-9)$ is $y=5x-19$ .

(c)

To determine

To graph: The function $f(x)=4x^2-11x-3$ and the tangent line.

Explanation of Solution

Given Information:

The function, $f(x)=4x^2-11x-3$.

Graph:

To graph the function $f(x)=4x^2-11x-3$,

Find $y-\text{intercept}$, $x-\text{intercept}$ and vertex.

To find $y-\text{intercept}$ set $x=0$.

$\Rightarrow f(0)=4(0)-11(0)-3=-3$

Thus, the $y-\text{intercept}$ is $(0,-3)$.

To find $x-\text{intercepts}$ set $f(x)=0$.

$\Rightarrow 0=4x^2-11x-3$

$\Rightarrow 0=(4x+1)(x-3)$

$\Rightarrow x=-\frac{1}{4},3$

Thus, the $x-\text{intercepts}$ are $\left(-\frac{1}{4},0\right)$ and $\left(3,0\right)$.

To find vertex of parabola, Compute $x=-\frac{b}{2a}$, where $a=4\text{and}b=-11$

$\Rightarrow x=-\frac{(-11)}{2\times 4}$

$\Rightarrow x=\frac{11}{8}=1.38$

Substitute $x$ value in $f(x)=4x^2-11x-3$ to find corresponding $y$ value,

$f\left(\frac{11}{8}\right)=4{\left(\frac{11}{8}\right)}^2-11\left(\frac{11}{8}\right)-3=-10.563$

Thus, the vertex is $\left(1.38,-10.563\right)$.

Graph of the function using $y-\text{intercept}$, $x-\text{intercepts}$ and vertex as follows:

From part (b),

The equation of tangent line to the graph of $f(x)=4x^2-11x-3$ at the point $(2,-9)$ is $y=5x-19$.

One point on the tangent line is $(2,-9)$.

For other point, substitute $x=3$ in $y=5x-19$.

$y=5\left(3\right)-19=15-19=-4$.

Thus, point is $(3,-4)$.

Plot these two points and draw the tangent line passing through them.

The graph is as below,

Interpretation:

The graph represents the equation of function $f(x)=4x^2-11x-3$ and the equation of tangent line $y=5x-19$ at point $(2,-9)$ .