Question #179868

I. Investigate whether the following functions are even or odd:

(a) f(x) = x3

(b) f(x) = cos x


II. State the mean value theorem

III. (a) Find the derivative of the function y = 2x2 + 12/x2 when x = 2

(b) f(x) = -3/x-7. Find the inverse of the function.

Iv. Consider the function f(x) = erx Determine the values of r so that f satisfies the equation f"(x) + f'(x) - 6f(x) = 0.





Expert's answer

  1. The function f(x)f(x) is called odd if and only if −f(x)=f(−x).-f(x)=f(-x). It is called even if and only if f(x)=−f(x)f(x)=-f(x).

(a) The function is odd, since −x3=(−x)3-x^3=(-x)^3 .

(b) The function is even, since cos(x)=cos(−x)cos(x)=cos(-x) .

2. The mean value Theorem states the following: Suppose that function f(x)f(x) is continuous on the closed interval [a,b][a,b] and is differentiable on the open interval (a,b)(a,b). Then there exists the point c∈(a,b)c\in(a,b) such that f′(c)=f(b)−f(a)b−af'(c)=\frac{f(b)-f(a)}{b-a}.

3. (a) The derivative of the function is: y′=4x−24x−3y'=4x-24x^{-3}. After substituting x=2x=2 we receive: y′(2)=8−248=8−3=5y'(2)=8-\frac{24}{8}=8-3=5.

(b) We receive: f(x)+7=−3xf(x)+7=-\frac3x. It yields: x=−3f(x)+7x=-\frac{3}{f(x)+7}. It can be also rewritten as: f−1(x)=−3x+7f^{-1}(x)=-\frac{3}{x+7}.

4. We substitute f(x)=erxf(x)=e^{rx} and receive: r2+r−6=0r^2+r-6=0 ; r=−1±52=2;−3.r=\frac{-1\pm5}{2}=2;-3. The solutions are: f(x)=e2xf(x)=e^{2x} and f(x)=e−3xf(x)=e^{-3x}.


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