Find the equation of the line tangent to the graph of the given function at the point $x = 1$ . $f ( x ) = \left( x ^ { 3 } + 3 \right) \left( x ^ { 2 } + x \right)$
A) $y = 8$
B) $y = 18 x - 10$
C) $y = 8 x$
D) $y = 18$
E) $y = 18 x$

Question

Find the equation of the line tangent to the graph of the given function at the point $x = 1$ .   $f ( x ) = \left( x ^ { 3 } + 3 \right) \left( x ^ { 2 } + x \right)$  
A) $y = 8$
B) $y = 18 x - 10$
C) $y = 8 x$
D) $y = 18$
E) $y = 18 x$

Quizplus · Accepted Answer

To find the equation of the tangent line at $x = 1$, we first need to find the derivative of $f(x)$ to get the slope of the tangent line at that point. The function $f(x) = (x^3 + 3)(x^2 + x)$ can be simplified before differentiation or directly differentiated using the product rule. Using the product rule, $f'(x) = (3x^2)(x^2 + x) + (x^3 + 3)(2x + 1)$. Evaluating $f'(1)$ gives the slope of the tangent line at $x = 1$. First, find $f'(1)$:$$f'(x) = 3x^2(x^2 + x) + (x^3 + 3)(2x + 1)$$$$f'(1) = 3(1)^2(1^2 + 1) + (1^3 + 3)(2(1) + 1)$$$$f'(1) = 3(2) + (4)(3)$$$$f'(1) = 6 + 12 = 18$$So, the slope of the tangent line at $x = 1$ is 18.Next, find the y-coordinate of the point on the graph at $x = 1$:$$f(1) = (1^3 + 3)(1^2 + 1) = (4)(2) = 8$$The point is $ (1, 8) $.Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is a point on the line, we get:$$y - 8 = 18(x - 1)$$$$y = 18x - 18 + 8$$$$y = 18x - 10$$Therefore, the equation of the tangent line at $x = 1$ is $y = 18x - 10$, which corresponds to choice B.

Find the Equation of the Line Tangent to the Graph x=1x = 1x=1

Find the Equation of the Line Tangent to the Graph $x = 1$