OMPT applications of derivatives practice: lesson and 10 problems

Answer: 1

I set $f^{\prime }(x)=3x^2-12x+9=3(x-1)(x-3)=0$, giving critical points $x=1$ and $x=3$. The derivative changes from positive to negative at $x=1$ (and negative to positive at $x=3$), so the local maximum sits at $x=1$. The second derivative confirms it: $f^{\prime \prime }(1)=6-12=-6<0$.

Answer: A — $y=6x-9$

The slope is $y^{\prime }=2x=6$ at $x=3$, and the point of tangency is $(3,9)$. Point-slope form gives $y-9=6(x-3)$, so $y=6x-9$. Checking: at $x=3$ the line gives $18-9=9$, matching the curve.

Answer: 10

Differentiating: $S^{\prime }(x)=1-\frac{100}{x^2}$, which vanishes when $x^2=100$, so $x=10$ (the negative root is outside the domain). Since $S^{\prime \prime }(x)=\frac{200}{x^3}>0$ there, this is indeed a minimum, with value $S(10)=20$.

Answer: B — $(-1,1)$

Decreasing means $f^{\prime }(x)<0$. Here $f^{\prime }(x)=3x^2-3=3(x-1)(x+1)$, which is negative exactly between the roots, on $(-1,1)$. Outside that interval the parabola $3x^2-3$ is positive and $f$ climbs.

Answer: Step 3

Steps 1 and 2 are sound. Step 3 misreads the second-derivative test: $f^{\prime \prime }(0)=0$ is inconclusive, not evidence of a minimum. Checking the sign of $f^{\prime }(x)=4x^2(x-3)$ around $x=0$: it is negative on both sides (the $x^2$ factor never changes sign), so $f$ just keeps decreasing through $x=0$ — neither a max nor a min.

Answer: 19.6

The velocity is $h^{\prime }(t)=-9.8t+19.6$, which is zero at $t=2$ seconds — the top of the flight. Substituting back: $h(2)=-4.9\cdot 4+19.6\cdot 2=-19.6+39.2=19.6$ metres.

Answer: A — $y=2x-8$

Slope ledger: $f^{\prime }(x)=2x-4$, so $f^{\prime }(3)=2$. Height ledger: $f(3)=9-12+1=-2$, so the point of contact is $(3,-2)$. The line $y=2x+b$ through that point needs $-2=6+b$, so $b=-8$ and the tangent is $y=2x-8$. Check: at $x=3$ the line gives $6-8=-2$, matching the curve.

Answer: -11

I solve $f^{\prime }(x)=3x^2-12=0$: $x^2=4$, so the candidates are $x=-2$ and $x=2$. Sign chart: $f^{\prime }(0)=-12<0$ and $f^{\prime }(3)=15>0$, so the derivative goes from negative to positive at $x=2$, a minimum. The question wants the value, not the location: $f(2)=8-24+5=-11$. Check against the other candidate: $f(-2)=-8+24+5=21$, the local maximum, and $21>-11$ as it should be.

Answer: 900

Let $x$ be the length of each side perpendicular to the wall and $y$ the length parallel to it. Fencing used: two sides of length $x$, the far side of length $y$, and the divider of length $y$, so $2x+2y=120$ and $y=60-x$. Area: $A(x)=x(60-x)=60x-x^2$ with $0<x<60$. Then $A^{\prime }(x)=60-2x=0$ gives $x=30$, and $A^{\prime }$ is positive before 30 and negative after, so it is a maximum. $A(30)=30\cdot 30=900$. Check the fence: $2\cdot 30+2\cdot 30=120$ metres.

Answer: 13

The derivative is $f^{\prime }(x)=3x^2-2kx+12$, a quadratic with discriminant $D=4k^2-144$. A local extreme needs $f^{\prime }$ to change sign, which requires two distinct roots, so no extremes means $D\le 0$: $k^2\le 36$, so $-6\le k\le 6$. The boundary values count: at $k=6$, $f^{\prime }(x)=3(x-2{)}^2\ge 0$, so $f$ is increasing throughout and has no extreme. The integers from $-6$ to $6$ inclusive number $13$. Check just outside: $k=7$ gives $D=196-144=52>0$, two extremes.