To find the tangent line to the curve y = x 3 + x 2 + 1 at the point ( 1 , 3 ) , we calculate the derivative to find the slope, which is 5 at this point. Using the point-slope form of a line, the equation of the tangent line is y = 5 x − 2 .
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Find the derivative of the function: d x d y = 3 x 2 + 2 x .
Evaluate the derivative at x = 1 to find the slope: m = 5 .
Use the point-slope form of a line: y − 3 = 5 ( x − 1 ) .
Simplify to find the equation of the tangent line: y = 5 x − 2 .
Explanation
Problem Analysis We are given the curve y = x 3 + x 2 + 1 and the point ( 1 , 3 ) . Our goal is to find the equation of the tangent line to the curve at this point.
Finding the Derivative First, we need to find the derivative of the function y = x 3 + x 2 + 1 with respect to x . This will give us the slope of the tangent line at any point on the curve. Using the power rule, we find the derivative to be: d x d y = 3 x 2 + 2 x
Evaluating the Derivative at x=1 Next, we need to evaluate the derivative at x = 1 to find the slope of the tangent line at the point ( 1 , 3 ) .
m = d x d y ∣ x = 1 = 3 ( 1 ) 2 + 2 ( 1 ) = 3 + 2 = 5 So, the slope of the tangent line at the point ( 1 , 3 ) is 5 .
Finding the Tangent Line Equation Now, we use the point-slope form of a line, which is given by y − y 1 = m ( x − x 1 ) , where ( x 1 , y 1 ) = ( 1 , 3 ) and m = 5 . Plugging in these values, we get: y − 3 = 5 ( x − 1 ) y − 3 = 5 x − 5 y = 5 x − 5 + 3 y = 5 x − 2
Final Answer Therefore, the equation of the tangent line to the curve y = x 3 + x 2 + 1 at the point ( 1 , 3 ) is y = 5 x − 2 .
Examples
Understanding tangent lines is crucial in various fields. For instance, in physics, the tangent line to a position-time curve gives the instantaneous velocity of an object. In economics, it can represent the marginal cost or revenue at a particular production level. Moreover, tangent lines are fundamental in optimization problems, helping to find maximum and minimum values of functions, which are essential in engineering and business applications.