What is the equation of the line tangent to #f(x)= -3x^5-8x^3+4x^2 # at #x=-1#?

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What is the equation of the line tangent to #f(x)= -3x^5-8x^3+4x^2 # at #x=-1#?

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Answer 1 · 2018-04-30T00:30:19

#31x+y+38=0#

Explanation:

#f(x)= -3x^5-8x^3+4x^2 implies#

#f^'(x)= -15x^4-24x^2+8x#

We have #f(-1) = -7# and #f^'(-1) = -31#. The tangent that we are seeking is thus a straight line having slope of #-31# going through #(-1,-7)#. So, its equation is

#y-(-7) = -31(x-(-1)) implies#

#31x+y+38=0#

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What is the equation of the line tangent to #f(x)= -3x^5-8x^3+4x^2 # at #x=-1#?

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