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I am new in differential equation. I just want to solve the following problem, Can any one give me the step wise solution for this question??? Determine the equation of the tangent line for the function f(x) = x2 + 1 at point (3,10).
I am new in differential equation. I just want to solve the following problem,
Can any one give me the step wise solution for this question???
Determine the equation of the tangent line for the function
f(x) = x2 + 1 at point (3,10).
Find the slope of the function by differentiation f '(x) = 2x Plug in the certain point's values Since this function does not have y we don't plug in y yet f '(3) = 6 {6 is now the slope of the point 3,10} Plug both slope and point values into a linear equation (y - y1) = m(x - x1) {this is the linear equation} (y - 10) = 6(x - 3) {Which can be simplified as below} y = 6x -8 Just as we can find the slope and equation of a tangent line for a function, we can also do the same for a normal line. However, the normal line has two differences from the tangent line. 1. The slope of a normal line is perpendicular to the slope of the tangent line. Or in other words, the negative inverse of the tangent line. 2. The normal line is only defined if x does not = 0. As a result, to find the slope and equation of the normal line, follow the steps above and convert the slope of the tangent line to the slope of the normal line.
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