Maxima AND Minima flash cards

1. 1.For a curve $y=f(x)$, write the equations of the tangent and normal at point $(x_1,y_1)$.

   Tangent: $y-y_1=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}(x-x_1)$; Normal: $y-y_1=-\dfrac{1}{\left(\frac{dy}{dx}\right)_{(x_1,y_1)}}(x-x_1)$. (If $\left(\frac{dy}{dx}\right)_{(x_1,y_1)}=0$, tangent is $y=y_1$ and normal is $x=x_1$.)

   Hint: Normal slope is the negative reciprocal of tangent slope.

2. 2.State the condition for a function $f(x)$ to be strictly increasing or decreasing on an interval $(a,b)$.

   $f$ is strictly increasing on $(a,b)$ if $f'(x)>0\ \forall x\in(a,b)$, and strictly decreasing if $f'(x)<0\ \forall x\in(a,b)$.

   Hint: Sign of the first derivative.

3. 3.Give the First Derivative Test for local maxima and minima at a critical point $x=c$ where $f'(c)=0$.

   If $f'(x)$ changes sign from positive to negative as $x$ increases through $c$, then $x=c$ is a point of local maximum; if it changes from negative to positive, $x=c$ is a point of local minimum; if $f'(x)$ does not change sign, $x=c$ is neither a maximum nor a minimum (a point of inflection).

   Hint: Check the sign of $f'(x)$ just before and after $c$.

4. 4.State the Second Derivative Test for extrema at a critical point $x=c$.

   If $f'(c)=0$ and $f''(c)<0$, then $x=c$ is a point of local maximum; if $f'(c)=0$ and $f''(c)>0$, then $x=c$ is a point of local minimum; if $f'(c)=0$ and $f''(c)=0$, the test fails and the First Derivative Test must be used.

   Hint: Uses the sign of $f''(c)$ at a point where $f'(c)=0$.

5. 5.State Rolle's Theorem and the Mean Value Theorem (Lagrange's) with their conditions.

   Rolle's Theorem: If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists $c\in(a,b)$ such that $f'(c)=0$.
   Lagrange's Mean Value Theorem: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in(a,b)$ such that $f'(c)=\dfrac{f(b)-f(a)}{b-a}$.

   Hint: Rolle's is the special case of LMVT when $f(a)=f(b)$.

6. 6.Find the absolute maximum and minimum of $f(x)=x^3-3x^2-9x+5$ on $[-2,4]$.

   $f'(x)=3x^2-6x-9=3(x-3)(x+1)$, giving critical points $x=-1,3$, both inside $[-2,4]$. Evaluating: $f(-2)=3$, $f(-1)=10$, $f(3)=-22$, $f(4)=-15$. So the absolute maximum is $10$ at $x=-1$ and the absolute minimum is $-22$ at $x=3$.

   Hint: Locate all critical points first, then compare with both endpoints.

7. 7.Find the absolute maximum and minimum of $f(x)=2x^3-15x^2+36x-48$ on $[0,4]$.

   $f'(x)=6x^2-30x+36=6(x-2)(x-3)$, so $x=2,3$ are critical points. Values: $f(0)=-48$, $f(2)=-20$, $f(3)=-21$, $f(4)=-16$. Although $x=2$ is a local max and $x=3$ a local min, the absolute maximum is $-16$ at the endpoint $x=4$ and the absolute minimum is $-48$ at the endpoint $x=0$ — the endpoints beat both interior extrema.

   Hint: Don't assume the interior stationary points automatically give the global values.

8. 8.Find the absolute maximum and minimum of $f(x)=x^4-4x^3+4x^2+1$ on $[-1,3]$.

   $f'(x)=4x^3-12x^2+8x=4x(x-1)(x-2)$, giving critical points $0,1,2$. Values: $f(-1)=10$, $f(0)=1$, $f(1)=2$, $f(2)=1$, $f(3)=10$. The absolute maximum $10$ is attained at both $x=-1$ and $x=3$, and the absolute minimum $1$ is attained at both $x=0$ and $x=2$ — a case where each global extreme value occurs at two distinct points.

   Hint: Compute all critical values and both endpoints; watch for repeated extreme values.

9. 9.Find the absolute maximum and minimum values of $f(x)=\sin x+\cos 2x$ on $[0,2\pi]$.

   Write $\cos2x=1-2\sin^2x$, so $f=1+\sin x-2\sin^2x$. Let $t=\sin x\in[-1,1]$ and $g(t)=-2t^2+t+1$; $g'(t)=1-4t=0\Rightarrow t=\tfrac14$, giving $g(\tfrac14)=\tfrac98$. Comparing with the endpoints $g(-1)=-2$ and $g(1)=0$, the absolute maximum is $\tfrac98$ (at $\sin x=\tfrac14$) and the absolute minimum is $-2$ (at $x=\tfrac{3\pi}{2}$).

   Hint: Substitute $t=\sin x$ to reduce to a quadratic on a closed interval.

10. 10.Find the absolute maximum and minimum of $f(x)=x-\ln x$ on $\left[\tfrac12,e\right]$.

    $f'(x)=1-\tfrac1x=0\Rightarrow x=1$, and $f''(x)=1/x^2>0$ so $x=1$ is a local minimum with $f(1)=1$. Comparing endpoints: $f(\tfrac12)=\tfrac12+\ln2\approx1.193$ and $f(e)=e-1\approx1.718$. So the absolute minimum is $1$ at $x=1$ and the absolute maximum is $e-1$ at $x=e$.

    Hint: Differentiate, find the single critical point, then compare with both ends.

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