Is #f(x)=(x+2)^2-4x^2-3x # increasing or decreasing at #x=-1 #? Calculus Graphing with the First Derivative Interpreting the Sign of the First Derivative (Increasing and Decreasing Functions) 1 Answer Özgür Özer Dec 7, 2015 #f'(-1)# is increasing at #x=-1# Explanation: #color(white)(xx)f(x)=(x+2)^2-4x^2-3x# #=>f(x)=x^2+4x+4-4x^2-3x# #=>f(x)=-3x^2+x+4# #=>f'(x)=-6x+1# #=>f'(-1)=6+1# #=>f'(-1)=7# #=>f'(-1)>0# so #f'(-1)# is increasing at #x=-1# Answer link Related questions How do you know a function is increasing? What is the meaning of monotonically increasing function? How do you find all intervals where the function #f(x)=1/3x^3+3/2x^2+2# is increasing? How do you find all intervals where the function #f(x)=e^(x^2)# is increasing? What is the derivative graph of a parabola? If #f(x) = 2x^3 + 3x^2 - 180x#, how do I find the intervals on which f is increasing and decreasing? Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) =... What information does the first derivative tell you? How do you find the interval in which the function #f(x)=2x^3 + 3x^2+180x# is increasing or decreasing? How do you find values of t in which the speed of the particle is increasing if he position of a... See all questions in Interpreting the Sign of the First Derivative (Increasing and Decreasing Functions) Impact of this question 1132 views around the world You can reuse this answer Creative Commons License