Is #f(x)=4x^5-x^4-9x^3+5x^2-7x# concave or convex at #x=-1#?
1 Answer
May 11, 2016
concave at x = -1
Explanation:
To determine if a function is concave/convex we calculate the value of f''(x) at x = a.
• If f''(a) > 0 , then f(x) is convex at x = a
• If f''(a) < 0 , then f(x) is concave at x = a
#f(x)=4x^5-x^4-9x^3+5x^2-7x# differentiate using the
#color(blue)" power rule"#
#rArrf'(x)=20x^4-4x^3-27x^2+10x-7# and
#f''(x)=80x^3-12x^2-54x+10# so
#f''(-1)=80(-1)^3-12(-1)^2-54(-1)+10#
#=-80-12+54+10=-28# Since f''(-1) < 0 , then f(x) is concave at x = -1
graph{4x^5-x^4-9x^3+5x^2-7x [-36.52, 36.53, -18.24, 18.29]}