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

1 Answer
Aug 2, 2018

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]}