Question

1. find all points (x,y) on the graph of f(x)=2x^3 - x + 4 with tangent lines parallel to the line y = 5x + 3.

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=2x^{3}-x + 4$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=6x^{2}-1$.

Step2: Determine the slope of the given line

The line $y = 5x+3$ has a slope $m = 5$.

Step3: Set the derivative equal to the slope

Since the tangent lines are parallel to $y = 5x + 3$, we set $f'(x)=5$. So, $6x^{2}-1=5$.

Step4: Solve the equation for $x$

Add 1 to both sides of the equation: $6x^{2}=6$. Then divide both sides by 6 to get $x^{2}=1$. Taking the square root of both sides, we have $x=\pm1$.

Step5: Find the corresponding $y$ - values

When $x = 1$, $y=f(1)=2(1)^{3}-1 + 4=2 - 1+4=5$.
When $x=-1$, $y=f(-1)=2(-1)^{3}-(-1)+4=-2 + 1+4=3$.

Answer:

$(1,5)$ and $(-1,3)$