Question

question
find the equation of the line tangent to the graph of (y = f(x)=x^{3}+4x + 2) at (x = 1).
provide your answer below:

Explanation:

Step1: Find the derivative of the function

The function is $f(x)=x^{3}+4x + 2$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=3x^{2}+4$.

Step2: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into $f'(x)$. Then $f'(1)=3(1)^{2}+4=3 + 4=7$. The slope of the tangent line $m = 7$.

Step3: Find the function value at $x = 1$

Substitute $x = 1$ into $f(x)$. So $f(1)=1^{3}+4\times1 + 2=1 + 4+2=7$. The point on the curve is $(1,7)$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(1,7)$ and $m = 7$. Substituting these values, we get $y - 7=7(x - 1)$.

Step5: Simplify the equation

Expand the right - hand side: $y - 7=7x-7$. Then $y=7x$.

Answer:

$y = 7x$