007A Sample Midterm 1, Problem 3 Detailed Solution

Let  $y=2x^{2}-3x+1.$

(a) Use the definition of the derivative to compute   ${\frac {dy}{dx}}.$

(b) Find the equation of the tangent line to  $y=2x^{2}-3x+1$  at  $(2,3).$

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Background Information:
Recall
$f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}$

Solution:

(a)

Step 1:
Let  $f(x)=2x^{2}-3x+1.$
Using the limit definition of the derivative, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {2(x+h)^{2}-3(x+h)+1-(2x^{2}-3x+1)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {2x^{2}+4xh+2h^{2}-3x-3h+1-2x^{2}-3x-1}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {4xh+2h^{2}-3h}{h}}.}\end{array}}$

Step 2:
Now, we simplify to get
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {h(4x+2h-3)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}4x+2h-3}\\&&\\&=&\displaystyle {4x-3.}\end{array}}$

(b)

Step 1:
We start by finding the slope of the tangent line to  $f(x)=2x^{2}-3x+1$  at  $(2,3).$
Using the derivative calculated in part (a), the slope is
${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {f'(2)}\\&&\\&=&\displaystyle {4(2)-3}\\&&\\&=&\displaystyle {5.}\end{array}}$

Step 2:
Now, the tangent line to  $f(x)=2x^{2}-3x+1$  at  $(2,3)$
has slope  $m=5$  and passes through the point  $(2,3).$
Hence, the equation of this line is
$y=5(x-2)+3.$
If we simplify, we get
$y=5x-7.$

Final Answer:
(a)     $4x-3$
(b)     $y=5x-7$

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