What are the extrema of #f(x)=5x^2+4x-3# on #[-oo,oo]#?

1 Answer
Aug 2, 2018

Answer:

Local minimum: #(-2/5, -19/5)#

Explanation:

Local extrema occur when the first derivative returns a zero whereas the second derivative returns a non-zero value.

Apply the Power Rule to find the first and second derivative of #f(x)#

#f'(x) = (2 xx 5) color(white)(l)x + 4 = 10 color(white)(l)x + 4#
#f''(x) = 10#

Set #f'(x)# to zero and solve for #x#:

#10 color(white)(l) x + 4 = f'(x) = 0#
#x = -2/5#

The solution to #f'(x) = 0# is unique, meaning that #f(x)# has only a single local extremum on #x in (-infty, +infty)#.

Evaluate the second derivative at #x = -2/5#:

#f''(-2/5) = 10 > 0#

#f'(x)= 0# and #f''(x)# is positive at #x =-2/5#. Therefore #f(x)# is concave upward and has a local minimum with coordinates #(-2/5, f(-2/5))#.

#f(-2/5) = -19/5 = -3.8#. Therefore #f(x)# has a local minimum at #(-2/5, -19/5)# or equivalently #(-0.4, -3.8)# as seen on the plot:

graph{5 x ^2 + 4x -3 [-2, 1, -5.567, 0.676]}