Question

write an equation of the line tangent to the graph of ( f(x) = 2x^{3}-x^{2}+3x ) at the point where ( x = 1 ).

Explanation:

Step1: Find the y - coordinate of the point

Substitute \(x = 1\) into \(f(x)=2x^{3}-x^{2}+3x\).
\(f(1)=2(1)^{3}-(1)^{2}+3(1)=2 - 1+3=4\). So the point is \((1,4)\).

Step2: Find the derivative of \(f(x)\)

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(f^\prime(x)=(2x^{3}-x^{2}+3x)^\prime=6x^{2}-2x + 3\).

Step3: Find the slope of the tangent line

Substitute \(x = 1\) into \(f^\prime(x)\). \(f^\prime(1)=6(1)^{2}-2(1)+3=6 - 2+3=7\). The slope \(m = 7\).

Step4: Use the point - slope form \(y - y_{1}=m(x - x_{1})\)

Here \(x_{1}=1,y_{1}=4,m = 7\). So \(y - 4=7(x - 1)\).
Simplify: \(y-4=7x-7\), then \(y=7x - 3\).

Answer:

\(y = 7x-3\)