Question

Need help on both of these Calculus AB mock exams.

Answer 1

AB1

(a) $g$ is continuous at $x=3$ if the two-sided limit exists. By definition of $g$, we have $g(3)=f(3^2-5)=f(4)$. We need to have $f(4)=1$, since continuity means

$\displaystyle\lim_{x\to3^-}g(x)=\lim_{x\to3}f(x^2-5)=f(4)$

$\displaystyle\lim_{x\to3^+}g(x)=\lim_{x\to3}5-2x=-1$

By the fundamental theorem of calculus (FTC), we have

$f(6)=f(4)+\displaystyle\int_4^6f'(x)\,\mathrm dx$

We know $f(6)=-4$, and the graph tells us the *signed* area under the curve from 4 to 6 is -3. So we have

$-4=f(4)-3\implies f(4)=-1$

and hence $g$ is continuous at $x=3$.

(b) Judging by the graph of $f'$, we know $f$ has local extrema when $x=0,4,6$. In particular, there is a local maximum when $x=4$, and from part (a) we know $f(4)=-1$.

For $x>6$, we see that $f'\ge0$, meaning $f$ is strictly non-decreasing on (6, 10). By the FTC, we find

$f(10)=f(6)+\displaystyle\int_6^{10}f'(x)\,\mathrm dx=7$

which is larger than -1, so $f$ attains an absolute maximum at the point (10, 7).

(c) Directly substituting $x=3$ into $k(x)$ yields

$k(3)=\dfrac{\displaystyle3\int_4^9f'(t)\,\mathrm dt-12}{3e^{2f(3)+5}-3}$

From the graph of $f'$, we know the value of the integral is -3 + 7 = 4, so the numerator reduces to 0. In order to apply L'Hopital's rule, we need to have the denominator also reduce to 0. This happens if

$3e^{2f(3)+5}-3=0\implies e^{2f(3)+5}=1\implies 2f(3)+5=\ln1=0\implies f(3)=-\dfrac52$

Next, applying L'Hopital's rule to the limit gives

$\displaystyle\lim_{x\to3}k(x)=\lim_{x\to3}\frac{9f'(3x)-4}{6f'(x)e^{2f(x)+5}-1}=\frac{9f'(9)-4}{6f'(3)e^{2f(3)+5}-1}=-\dfrac4{21e^{2f(3)+5}-1}$

which follows from the fact that $f'(9)=0$ and $f'(3)=3.5$.

Then using the value of $f(3)$ we found earlier, we end up with

$\displaystyle\lim_{x\to3}k(x)=-\dfrac4{20}=-\dfrac15$

(d) Differentiate both sides of the given differential equation to get

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=2f''(2x)(3y-1)+f'(2x)\left(3\dfrac{\mathrm dy}{\mathrm dx}\right)=[2f''(2x)+3f'(2x)^2](3y-1)$

The concavity of the graph of $y=h(x)$ at the point (1, 2) is determined by the sign of the second derivative. For $y>\frac13$, we have $3y-1>0$, so the sign is entirely determined by $2f''(2x)+3f'(2x)^2$.

When $x=1$, this is equal to

$2f''(2)+3f'(2)^2$

We know $f''(2)=0$, since $f'$ has a horizontal tangent at $x=2$, and from the graph of $f'$ we also know $f'(2)=5$. So we find $h''(1)=75>0$, which means $h$ is concave upward at the point (1, 2).

(e) $h$ is increasing when $h'>0$. The sign of $h'$ is determined by $f'(2x)$. We're interested in the interval $-1$, and the behavior of $h$ on this interval depends on the behavior of $f'$ on $-2$.

From the graph, we know $f'>0$ on (0, 4), (6, 9), and (9, 10). This means $h'>0$ on (0, 2), (3, 4.5), and (4.5, 5).

(f) Separating variables yields

$\dfrac{\mathrm dy}{3y-1}=f'(2x)\,\mathrm dx$

Integrating both sides gives

$\dfrac13\ln|3y-1|=\dfrac12f(2x)+C\implies\ln|3y-1|=\dfrac32f(2x)+C$