At what value(s) of x does f(x)=x^4-18x^2 have a critical point where the graph changes from decreasing to increasing?
take f"(x)=0 to find critical values v1,v2,...
take f (vn)
produces inflection points
Answer:
\(x = 2, x = -2\) are the critical points where the graph changes from decreasing to increasing.
Step-by-step explanation:
We are given the expression:
\(f(x) = x^4 - 18x^2\)
We have to find the critical points of the given expression.
First we differentiate the given expression with respect to x.
\(\displaystyle\frac{d(f(x))}{dx} = 4x^3 - 16x\)
Equating f'(x) to zero, to obtain the critical points:
\(f'(x) = 0\\4x^3 - 16x = 0\\x(4x^2 - 16) = 0\\x(2x-4)(2x+4) = 0\\x = 0, x = 2, x = -2\)
Hence, \(x = 2, x = -2\) are the critical points where the graph changes from decreasing to increasing.