Question

1. (#11 exam 1 review) find the equation of the line tangent to the graph of the function: $f(x)=x - 2x^{2}$ at $x = 1$. you must use the definition of the derivative to obtain the derivative.

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function $y = f(x)$ is $f^\prime(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=x - 2x^{2}$, then $f(x + h)=(x + h)-2(x + h)^{2}=(x + h)-2(x^{2}+2xh+h^{2})=x + h-2x^{2}-4xh - 2h^{2}$.

Step2: Calculate $f(x + h)-f(x)$

$f(x + h)-f(x)=(x + h-2x^{2}-4xh - 2h^{2})-(x - 2x^{2})=h-4xh - 2h^{2}$.

Step3: Find $\frac{f(x + h)-f(x)}{h}$

$\frac{f(x + h)-f(x)}{h}=\frac{h-4xh - 2h^{2}}{h}=1-4x - 2h$.

Step4: Calculate the limit as $h

ightarrow0$
$f^\prime(x)=\lim_{h
ightarrow0}(1-4x - 2h)=1-4x$.

Step5: Evaluate the derivative at $x = 1$

When $x = 1$, $f^\prime(1)=1-4\times1=-3$. This is the slope $m$ of the tangent - line.

Step6: Find the point on the function at $x = 1$

When $x = 1$, $f(1)=1-2\times1^{2}=-1$. So the point $(x_1,y_1)=(1,-1)$.

Step7: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$. Substituting $m=-3$, $x_1 = 1$, and $y_1=-1$ gives $y+1=-3(x - 1)$.

Step8: Simplify the equation

$y+1=-3x + 3$, so $y=-3x+2$.

Answer:

$y=-3x + 2$