Question #181106

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.




1
Expert's answer
2021-04-29T17:51:29-0400

I)

A function f(x) is said to be:

even, if f(-x) =f(x) and

odd, if f(-x)=-f(x)

Therefore, for

(a) f(x)=x3 , we have f(-x)=(-x)3 = (-1)3(x)3= -x3=-f(x)

Hence, f(x)=x3 is an odd function.



(b) f(x)=cos x,

so, f(-x)=cos (-x)=cos (0-x)

Using the identity, cos (A-B)=cos A cos B + sin A sin B, we have

f(-x)=cos (-x)=cos (0-x)=cos (0) cos (x) + sin (0) sin (x)

=cos (x) - 0 sin (x)

=cos (x)

=f(x)

Hence, cos (x) is an even function.


II) The mean value theorem is stated as follows:

Suppose g(x) is a function that is continuous on a closed interval [a, b] and

is also differentiable on the open interval (a, b) then there exists a number c

such that a<c<b and

g(c)=g(b)g(a)bag'(c) = \frac{g(b)-g(a)}{b-a}

III) a) y=2x2+12x2y=2x^2+\frac{12}{x^2}

y(x)=4x24x3y'(x) = 4x-\frac{24}{x^3}

when x=2, we have

y(2)=4(2)24(2)3=83=5y'(2)=4(2)-\frac{24}{(2)^3}=8-3=5


b) Let y=f(x)

So, y=3x7y=\frac{-3}{x}-7

swap x with y: x=3y7x=\frac{-3}{y}-7

solve for y: xy=37yxy=-3-7y

y(x+7)=3y(x+7)=-3

y=3x+7y=\frac{-3}{x+7}

Thus, f1(x)=3x+7f^{-1}(x) = \frac{-3}{x+7}


IV) From f(x)=erx,f(x)= e^{rx}, we have f(x)=rerxf'(x)= re^{rx} and f(x)=r2erxf''(x)= r^2e^{rx}

Substituting the derivatives into the equation f(x)+f(x)6f(x)=0,f''(x) + f'(x)-6f(x)=0, we have

r2erx+rerx6erx=0(r2+r6)erx=0r^2e^{rx}+re^{rx}-6e^{rx}=0\\ (r^2+r-6)e^{rx}=0

Since erx0,e^{rx}\neq 0, we have (r2+r6)=0(r^2+r-6)=0

Upon solving the quadratic equation, we have

r2+r6=(r+3)(r2)=0r^2+r-6=(r+3)(r-2)=0

So, r=2r=2 or r=3r=-3



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