f=x^2(e^x)
f' = x(x+2)e^x
Clearly, x=0 or x = -2
f(-2) = 4/e^2
f=(x^3+27)^(1/2)
f'= 3x^2/(2√(x^3+27))
the denominator is zero when x = -3
1. f(x)=x^2(e^x) find a point where the tangent is horizontal.
The answer is apparently (-2, 4/e^2). But I got 0 when I solved for the derivative. I don't get this.
2. What value of x does the function f(x)=(x^3+27)^(1/2) have vertical tangents?
My approach was to find undefined values, and so x basically couldn't be smaller than -27, yet the answer is -3. How?
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