Question

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) = e^rx Determine the values of r so that f satisfies the equation f"(x) + f'(x) - 6f(x) = 0.

Answer 1

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

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

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

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

when x=2, we have

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

b) Let y=f(x)

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

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

solve for y: $xy=-3-7y$

$y(x+7)=-3$

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

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

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

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

$r^2e^{rx}+re^{rx}-6e^{rx}=0\\ (r^2+r-6)e^{rx}=0$

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

Upon solving the quadratic equation, we have

$r^2+r-6=(r+3)(r-2)=0$

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