Tutor profile: Mark S.

Let $f(x) = x^2 - 8x + 15$ and $g(x) = 2x + 5$. Find the function compositions $(f \circ g)(x)$ and $(g \circ f)(x) $.

The function composition $(f \circ g)(x)$ is defined to be $f(g(x))$, where $g(x)$ is the input to the $f$ function. In this case, $g(x) = 2x + 5$, and so we plug in $2x+5$ into the $f$ function, replacing $x$: $$ (2x + 5)^2 - 8(2x+5) + 15 = (2x+5)(2x+5) - 8(2x) - 8(5) + 15 = 2x(2x) + 2x(5) + 5(2x) + 5(5) - 16x - 40 + 15 = 4x^2 + 10x + 10x + 25 - 16x - 25 = 4x^2 + 4x $$ The function composition $(g \circ f)(x)$ is defined to be $g(f(x))$, where $f(x)$ is the input to the $g$ function. In this case, $f(x) = x^2 - 8x + 15$, and so we plug in $x^2 - 8x + 15$ into the $g$ function, replacing $x$: $$ 2(x^2 - 8x + 15) + 5 = 2x^2 - 2(8x) + 2(15) + 5 = 2x^2 - 16x + 30 + 5 = 2x^2 - 16x + 35 $$ Hence we have $ (f \circ g)(x) = 4x^2 + 4x $ and $ (g \circ f)(x) = 2x^2 - 16x + 35 $.

Verify the trigonometric identity $ \cos^4 t - \sin^4 t = 1 - 2\sin^2 t $. When applicable, identify the trigonometric identity used in your verification steps.

To verify a trigonometric identity, we need to start with one side of the identity and perform algebraic manipulations or utilize other identities in order to arrive at the other side of the identity. We can begin with the left-hand side, with $ \cos^4 t - \sin^4 t $. This looks almost like one of the double-angle identities for cosine: $ \cos 2t = \cos^2 t - \sin^2 t $. But the powers are different, so we need to perform some algebra first: $$ \cos^4 t - \sin^4 t = \left( \cos^2 t \right)^2 - \left( \sin^2 t \right)^2 = \left( \cos^2 t - \sin^2 t \right) \left( \cos^2 t + \sin^2 t \right) $$ Note that we used the difference of squares to factor, after first rewriting both terms as the square of a square. Next, we note that the latter factor $ \cos^2 t + \sin^2 t $ is a Pythagorean identity, which we know to be 1: $$ \left( \cos^2 t - \sin^2 t \right) \left( \cos^2 t + \sin^2 t \right) = \left( \cos^2 t - \sin^2 t \right) (1) = \cos^2 t - \sin^2 t $$ Now we have the previously noted double-angle identity for cosine: $$ \cos^2 t - \sin^2 t = \cos 2t $$ We are not yet immediately done as we have not yet yielded the right-hand side of the identity, but remember that cosine actually has three different formulas for its double-angle, one of which is precisely the right-hand expression: $$ \cos 2t = 1 - 2\sin^2 t $$ So we have successfully proven that $ \cos^4 t - \sin^4 t = 1 - 2 \sin^2 t$, as desired.

Determine whether the Mean Value Theorem applies to the function $f(x) = x^2 - 8x + 15$ on the interval $[1,5]$, and if so, find a value of $c$ as guaranteed by the conclusion of the theorem.

The Mean Value Theorem states that, if a function $f(x)$ is continuous on the closed interval $[a,b]$ and is differentiable on the open interval $(a,b)$, then there is some point $c$ in the interval $(a,b)$ for which $f'(c) = \frac{f(b)-f(a)}{b-a}$. To begin, we must first determine that the hypotheses hold. Since $f(x)$ is a polynomial, it is therefore continuous (on the interval $[1,5]$), so the first hypothesis is true. To determine differentiability, we need to find $f'(x)$ and show that it is defined for all numbers in the interval $(1,5)$. A simple application of the Power Rule gives us $f'(x) = 2x - 8$, which is again a polynomial and is therefore defined for all numbers (and in particular on the interval $(1,5)$). Therefore, the hypotheses of the Mean Value Theorem hold, and so the MVT applies to $f(x)$. To find the value of $c$ as guaranteed by the theorem, we need to first find $\frac{f(5)-f(1)}{5-1}$: $$ \frac{f(5)-f(1)}{5-1} = \frac{(5^2 - 8*5 + 15) - (1^2 - 8*1 + 15)}{4} = \frac{0 - 8}{4} = \frac{-8}{4} = -2 $$ We thus need to find a $c$ that satisfies $f'(c) = -2$. But we know $f'(x) = 2x - 8$, and so $f'(c) = 2c - 8$; we need only solve the equation $ 2c - 8 = -2 $. Solving for $c$ gives $c = 3$. Thus the $c$ as guaranteed by the conclusion must be $c=3$.