EBUS CAL12 Chapter 4 Applications of Derivatives Test-1

Step 1: The first derivative \(f'(x)\) describes whether the function is increasing or decreasing. Step 2: Since \(f'(x)>0\), the function is increasing. Step 3: The second derivative \(f''(x)\) describes the concavity of the function. Step 4: Since \(f''(x)<0\), the graph is concave down. Step 5: Therefore, the function is increasing while bending downward. Answer: The function is increasing and concave down.

Step 1: The function \(f(x)=-4x^2-8x-16\) is a quadratic. Step 2: Since the leading coefficient \(-4\) is negative, the parabola opens downward, so its vertex gives the absolute maximum. Step 3: Use the vertex formula \(x=\frac{-b}{2a}\). Here \(a=-4\) and \(b=-8\). Step 4: Substitute: \(x=\frac{-(-8)}{2(-4)}=\frac{8}{-8}=-1\). Step 5: Evaluate the function at \(x=-1\): \(f(-1)=-4(-1)^2-8(-1)-16\). Step 6: Simplify: \(f(-1)=-4+8-16=-12\). Step 7: Therefore, the absolute maximum value is \(-12\). Answer: -12

Step 1: To find where \(f'(x)=0\), set the numerator equal to zero: \(x^2+8x-9=0\). Step 2: Factor the numerator: \(x^2+8x-9=(x+9)(x-1)\). Step 3: Set each factor equal to zero: \(x+9=0\) gives \(x=-9\), and \(x-1=0\) gives \(x=1\). Step 4: To find where \(f'(x)\) is undefined, set the denominator equal to zero: \(x+6=0\). Step 5: This gives \(x=-6\). Step 6: Add all values: \(-9+1+(-6)=-14\). Answer: -14

Step 1: A function is decreasing when \(d'(t)0\), the sign depends on \(-t^2+14t-1\). Step 5: Solve \(-t^2+14t-1<0\). This gives \(t7+4\sqrt{3}\). Because \(t>0\), decreasing occurs on \(0<t7+4\sqrt{3}\). Step 6: Concave up means \(d''(t)>0\). Differentiating again gives \(d''(t)=\frac{e^{-t/7}}{49}(t^2-28t+99)\). Step 7: Since \(e^{-t/7}>0\), solve \(t^2-28t+99>0\). Step 8: The roots are \(t=14\pm\sqrt{97}\), so concave up occurs when \(t14+\sqrt{97}\). Step 9: Intersect the decreasing intervals with the concave-up intervals. Step 10: The final intervals are \(0<t14+\sqrt{97}\). Answer: \(0<t14+\sqrt{97}\)

Step 1: The Mean Value Theorem says \(f'(c)=\frac{f(b)-f(a)}{b-a}\). Step 2: Here \(a=0\), \(b=10\), and \(f(x)=\sqrt{9x+7}\). Step 3: Compute the average rate of change: \(\frac{f(10)-f(0)}{10-0}=\frac{\sqrt{97}-\sqrt{7}}{10}\). Step 4: Find the derivative: \(f'(x)=\frac{9}{2\sqrt{9x+7}}\). Step 5: Set \(f'(c)=\frac{\sqrt{97}-\sqrt{7}}{10}\). Step 6: Solve \(\frac{9}{2\sqrt{9c+7}}=\frac{\sqrt{97}-\sqrt{7}}{10}\). Step 7: Solving gives \(c\approx3.56\), not \(0.46\). Step 8: Therefore, the statement is false. Answer: False

Step 1: Substitute \(x=0\) directly into the expression. Step 2: The numerator becomes \(\sin(3\cdot0)=\sin0=0\). Step 3: The denominator becomes \(e^{2\cdot0}-4=e^0-4=1-4=-3\). Step 4: Therefore, the limit is \(\frac{0}{-3}=0\). Step 5: The statement says the limit equals \(0\), so it is true. Answer: True

Step 1: The volume of a sphere is \(V=\frac{4}{3}\pi r^3\). Step 2: Differentiate both sides with respect to time: \(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\). Step 3: We are given \(\frac{dV}{dt}=130\) and \(r=8\). Step 4: Substitute into the formula: \(130=4\pi(8)^2\frac{dr}{dt}\). Step 5: Simplify: \(130=256\pi\frac{dr}{dt}\). Step 6: Solve for \(\frac{dr}{dt}\): \(\frac{dr}{dt}=\frac{130}{256\pi}\). Step 7: Calculate: \(\frac{130}{256\pi}\approx0.1616\). Step 8: Rounded to three significant figures, \(\frac{dr}{dt}\approx0.162\text{ cm/s}\). Answer: 0.162

Step 1: A point of inflection occurs when the concavity of a graph changes. Step 2: The condition \(f''(1)=0\) means the second derivative is zero at \(x=1\). Step 3: However, \(f''(1)=0\) alone does not guarantee a change in concavity. Step 4: The condition \(f'(1)=0\) only tells us that \(x=1\) may be a critical point. Step 5: To confirm an inflection point, we must check whether \(f''(x)\) changes sign around \(x=1\). Step 6: Since the statement says it must be an inflection point, the statement is false. Answer: False

Step 1: The area of the right triangle is \(A=\frac{1}{2}ab\). Step 2: We are given \(A=6\) and \(b=4\). Substitute: \(6=\frac{1}{2}a(4)\). Step 3: Simplify: \(6=2a\), so \(a=3\). Step 4: Use the Pythagorean theorem: \(a^2+b^2=c^2\). Step 5: Substitute \(a=3\) and \(b=4\): \(c^2=3^2+4^2=9+16=25\), so \(c=5\). Step 6: Differentiate \(a^2+b^2=c^2\) with respect to time: \(2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\). Step 7: Divide by \(2\): \(a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\). Step 8: We are given \(\frac{dc}{dt}=1\) and \(\frac{db}{dt}=3\frac{da}{dt}\). Step 9: Substitute \(a=3\), \(b=4\), \(c=5\): \(3\frac{da}{dt}+4\frac{db}{dt}=5(1)\). Step 10: Replace \(\frac{db}{dt}\) with \(3\frac{da}{dt}\): \(3\frac{da}{dt}+4(3\frac{da}{dt})=5\). Step 11: Simplify: \(15\frac{da}{dt}=5\), so \(\frac{da}{dt}=\frac{1}{3}\). Step 12: Therefore, \(\frac{db}{dt}=3\cdot\frac{1}{3}=1\). Answer: 1

Step 1: Let \(x\) be the distance from the pole to the player, and let \(s\) be the length of the player's shadow. Step 2: The distance from the pole to the tip of the shadow is \(x+s\). Step 3: Use similar triangles. The large triangle has height \(44\) and base \(x+s\). The small triangle has height \(5\) and base \(s\). Step 4: Set up the proportion: \(\frac{44}{x+s}=\frac{5}{s}\). Step 5: Cross multiply: \(44s=5(x+s)\). Step 6: Expand: \(44s=5x+5s\). Step 7: Simplify: \(39s=5x\). Step 8: Differentiate both sides with respect to time: \(39\frac{ds}{dt}=5\frac{dx}{dt}\). Step 9: The player walks away from the pole at \(\frac{dx}{dt}=13\). Step 10: Substitute: \(39\frac{ds}{dt}=5(13)=65\). Step 11: Solve: \(\frac{ds}{dt}=\frac{65}{39}=\frac{5}{3}\). Step 12: Therefore, the shadow length is increasing at \(\frac{5}{3}\text{ ft/sec}\). Answer: \(\frac{5}{3}\)