What is the equation of the normal line of #f(x)=x^3e^-x+x^3# at #x=-5#?

Question

Christopher Agresta · Accepted Answer

#y_(t)=color(red)m_tx+color(red)b_t#
The normal equation is #y_(n)= color(blue)m_nx+color(blue)b_n#
a) Where #color(red)m_t = -1/color(blue)m_n #
b) #color(red)b_t = color(blue)b_n # the slope of the normal line is the negative reciprocal of the slope of the tangent line. so what you need to don is : 1) find the the slope of f(x) at #x = -5 => f'(-5) # a) #f'(x)=3x^2e^-x - x^3e^-x + 3x^2 # b) #f'(-5)=y=75e^5 + 125e^5-125# 2) find the equation of line, equation of tangent at (-5, f(5)) a) Tangent equation, #y_(t)=color(red)m_tx+color(red)b_t# b) #color(red)m_t = f'(-5)# c) #y_(t)-f(-5)=color(red)f'(-5)(x+5)+color(red)b_t# solve for b 4) The normal equation is #y_(n)= color(blue)m_nx+color(blue)b_n# a) Where #color(red)m_t = -1/color(blue)m_n # b) #color(red)b_t = color(blue)b_n # There is some algebra that you need to, which I leave to you. the recipe for solving is there for you for now and future problems... Good luck

Christopher Agresta · Answer

undefined To find the equation of the normal line at a given point, we need to determine the slope of the tangent line at that point and then find the negative reciprocal of that slope.

To find the slope of the tangent line, we take the derivative of the function f(x).

The derivative of f(x) = x^3e^(-x) + x^3 can be found using the product rule and the chain rule.

The derivative is f'(x) = 3x^2e^(-x) - x^3e^(-x) + 3x^2.

To find the slope at x = -5, we substitute -5 into the derivative:

f'(-5) = 3(-5)^2e^(-(-5)) - (-5)^3e^(-(-5)) + 3(-5)^2.

Simplifying this expression gives us the slope of the tangent line at x = -5.

Once we have the slope, we can find the negative reciprocal to get the slope of the normal line.

Finally, we can use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point on the function, and m is the slope of the normal line.

Substituting the values, we can find the equation of the normal line.