Curve Sketching 2
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Question 1 of 5
1. Question
Sketch the curve`f(x)=3x^416x^3+24x^2+3`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA point that satisfies `f”(x)=0` is called an Inflection PointFirst, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `3x^416x^3+24x^2+3` `f'(x)` `=` `12x^348x^2+48x=0` Equate `f'(x)` to `0` `=` `12x(x^24x+4)=0` Factor out `12x` `=` `4x(x2)(x2)=0` Factor out further `x=0,x=2` Find the corresponding `y` values by substituting each `x` value to the function`f(x)` `=` `3x^416x^3+24x^2+3` $$f(\color{#007DDC}{0})$$ `=` $$3(\color{#007DDC}{0}^4)16(\color{#007DDC}{0}^3)+24(\color{#007DDC}{0}^2)+3$$ `=` `00+0+3` `=` `3` Stationary Point:
$$(\color{#007DDC}{0},\color{#e65021}{3})$$`f(x)` `=` `3x^416x^3+24x^2+3` $$f(\color{#007DDC}{2})$$ `=` $$3(\color{#007DDC}{2}^4)16(\color{#007DDC}{2}^3)+24(\color{#007DDC}{2}^2)+3$$ `=` `48128+96+3` `=` `19` Stationary Point:
$$(\color{#007DDC}{2},\color{#e65021}{19})$$Use the second derivative to check if each stationary point is either a maximum, minimum, or an inflection pointChecking $$(\color{#007DDC}{0},\color{#e65021}{3})$$:`f'(x)` `=` `12x^348x^2+48x` `f”(x)` `=` `36x^296x+48` $$f”(\color{#007DDC}{0})$$ `=` $$36(\color{#007DDC}{0}^2)96(\color{#007DDC}{0})+48$$ `=` `00+48` `=` `48``>``0` This means $$(\color{#007DDC}{0},\color{#e65021}{3})$$ is a minimum pointChecking $$(\color{#007DDC}{2},\color{#e65021}{19})$$:`f'(x)` `=` `12x^348x^2+48x` `f”(x)` `=` `36x^296x+48` $$f”(\color{#007DDC}{2})$$ `=` $$36(\color{#007DDC}{2}^2)96(\color{#007DDC}{2})+48$$ `=` `144192+48` `=` `0` This means $$(\color{#007DDC}{2},\color{#e65021}{19})$$ is an inflection pointIdentify where the curve decreases or increasesIt is clear that a minimum point would have a decreasing curve on the left and an increasing curve on the right. Hence, we just check the curves around the inflection pointStart by setting up a grid of points around `x=2`Test the gradient at `x=1``f'(x)` `=` `12x^348x^2+48x=0` $$f'(\color{#00880A}{1})$$ `=` $$12(\color{#00880A}{1})^348(\color{#00880A}{1}^2)+48(\color{#00880A}{1})$$ `=` `12(1)48(1)+48` `=` `1248+48` `=` `12` This value is positive, which means the curve’s slope at `1` is increasingIndicate this by adding an increasing slant line under `1`Test the gradient at `x=3``f'(x)` `=` `12x^348x^2+48x=0` $$f'(\color{#00880A}{3})$$ `=` $$12(\color{#00880A}{3})^348(\color{#00880A}{3}^2)+48(\color{#00880A}{3})$$ `=` `12(27)48(9)+144` `=` `324432+144` `=` `36` This value is positive, which means the curve’s slope at `3` is increasingIndicate this by adding an increasing slant line under `3`Finally, draw a curve along the stationary points, keeping in mind where it decreases or increases 
Question 2 of 5
2. Question
Sketch the curve`f(x)=(x^21)/(x^2+1)`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA point that satisfies `f”(x)=0` is called an Inflection PointFirst, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `(x^21)/(x^2+1)` `f'(x)` `=` `(4x)/((x^2+1)^2)=0` Equate `f'(x)` to `0` `=` `4x=0` Multiply both sides by `(x^2+1)^2` `=` `0=0` `x=0` will satisfy the equation `x=0` Find the corresponding `y` values by substituting each `x` value to the function`f(x)` `=` `(x^21)/(x^2+1)` $$f(\color{#007DDC}{0})$$ `=` $$\frac{\color{#007DDC}{0}^21}{\color{#007DDC}{0}^2+1}$$ `=` `1/1` `=` `1` Stationary Point:
$$(\color{#007DDC}{0},\color{#e65021}{1})$$Use the second derivative to check if the stationary point `(``0``,``1``)` is either a maximum, minimum, or an inflection point`f'(x)` `=` `(4x)/((x^2+1)^2)` `f”(x)` `=` `(4(x^2+1)^216x(x^2+1)^2)/((x^2+1)^4)` $$f”(\color{#007DDC}{0})$$ `=` $$\frac{4(\color{#007DDC}{0}^2+1)^216(\color{#007DDC}{0})(\color{#007DDC}{0}^2+1)^2}{(\color{#007DDC}{0}^2+1)^4}$$ `=` `4``>``0` This means $$(\color{#007DDC}{0},\color{#e65021}{1})$$ is a minimum pointNext, solve for the `y` and `x`intercepts of the curve by substituting `x=0` and `y=0` respectively to the formula`y`intercept`y` `=` `(x^21)/(x^2+1)` `y` `=` `(0^21)/(0^2+1)` Substitute `x=0` `y` `=` `1/1` `y` `=` `1` `x`intercept`y` `=` `(x^21)/(x^2+1)` `0` `=` `(x^21)/(x^2+1)` Substitute `y=0` `0` `=` `x^21` Multiply both sides by `x^2+1` `1` `=` `x^2` Add `1` to both sides `x` `=` `+1` Find the square root of both sides Next, apply the limit `∞` to `x` to check for asymptotes`lim_(x→∞)(x^21)/(x^2+1)` `=` `1` This means that the curve will be approaching but will not go beyond `y=1`Finally, draw a curve along the stationary points and intercepts, keeping in mind not to touch the asymptote 
Question 3 of 5
3. Question
Sketch the curve`f(x)=(x4)(x+2)^2`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA point that satisfies `f”(x)=0` is called an Inflection PointFirst, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `(x4)(x+2)^2` `f'(x)` `=` `3x^312=0` Equate `f'(x)` to `0` `=` `3(x^24)=0` Factor out `3` `=` `3(x+2)(x2)=0` Factor out further `x=2,x=2` Find the corresponding `y` values by substituting each `x` value to the function`f(x)` `=` `(x4)(x+2)^2` $$f(\color{#007DDC}{2})$$ `=` $$[(\color{#007DDC}{2})4][(\color{#007DDC}{2})+2]^2$$ `=` `(6)(0)` `=` `0` Stationary Point:
$$(\color{#007DDC}{2},\color{#e65021}{0})$$`f(x)` `=` `(x4)(x+2)^2` $$f(\color{#007DDC}{2})$$ `=` $$(\color{#007DDC}{2}4)(\color{#007DDC}{2}+2)^2$$ `=` `(2)(16)` `=` `32` Stationary Point:
$$(\color{#007DDC}{2},\color{#e65021}{32})$$Use the second derivative to check if each stationary point is either a maximum, minimum, or an inflection pointChecking $$(\color{#007DDC}{2},\color{#e65021}{0})$$:`f'(x)` `=` `3x^312` `f”(x)` `=` `6x` $$f”(\color{#007DDC}{2})$$ `=` $$6(\color{#007DDC}{2})$$ `=` `12``<``0` This means $$(\color{#007DDC}{2},\color{#e65021}{0})$$ is a maximum pointChecking $$(\color{#007DDC}{2},\color{#e65021}{32})$$:`f'(x)` `=` `12x^348x^2+48x` `f”(x)` `=` `3x^312` $$f”(\color{#007DDC}{2})$$ `=` $$6(\color{#007DDC}{2})$$ `=` `12``>``0` This means $$(\color{#007DDC}{2},\color{#e65021}{32})$$ is a minimum pointNext, substitute `x=0` to the original function to find the inflection point`f(x)` `=` `(x4)(x+2)^2` `f(0)` `=` `(04)(0+2)^2` Equate `x=0` `=` `(4)(4)` `=` `16` This means `(0,16)` is an inflection pointFinally, draw a curve along the stationary points, keeping in mind where it decreases or increases 
Question 4 of 5
4. Question
Sketch the curve`f(x)=x+1/x`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA point that satisfies `f”(x)=0` is called an Inflection PointFirst, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `x+1/x` `f'(x)` `=` `11/(x^2)=0` Equate `f'(x)` to `0` `=` `x^21=0` Multiply both sides by `x^2` `=` `x^2=1` Factor the difference of two squares `x=1,x=1` Find the corresponding `y` values by substituting each `x` value to the function`f(x)` `=` `x+1/x` $$f(\color{#007DDC}{1})$$ `=` $$\color{#007DDC}{1}+\frac{1}{\color{#007DDC}{1}}$$ `=` `1+1` `=` `2` Stationary Point:
$$(\color{#007DDC}{1},\color{#e65021}{2})$$`f(x)` `=` `x+1/x` $$f(\color{#007DDC}{1})$$ `=` $$\color{#007DDC}{1}+\frac{1}{\color{#007DDC}{1}}$$ `=` `11` `=` `2` Stationary Point:
$$(\color{#007DDC}{1},\color{#e65021}{2})$$Use the second derivative to check if each stationary point is either a maximum, minimum, or an inflection pointChecking $$(\color{#007DDC}{1},\color{#e65021}{2})$$:`f'(x)` `=` `11/(x^2)` `=` `1x^(2)` `f”(x)` `=` `2x^(3)` `=` `2/(x^3)` $$f”(\color{#007DDC}{1})$$ `=` $$\frac{2}{\color{#007DDC}{1}^3}$$ `=` `2``>``0` This means $$(\color{#007DDC}{1},\color{#e65021}{2})$$ is a minimum pointChecking $$(\color{#007DDC}{1},\color{#e65021}{2})$$:`f'(x)` `=` `11/(x^2)` `=` `1x^(2)` `f”(x)` `=` `2x^(3)` `=` `2/(x^3)` $$f”(\color{#007DDC}{1})$$ `=` $$\frac{2}{\color{#007DDC}{1}^3}$$ `=` `2``<``0` This means $$(\color{#007DDC}{1},\color{#e65021}{2})$$ is a maximum pointNext, apply the limits `0` and `∞` to `x` to check for asymptotes`x→0``lim_(x→0)1+1/x` `=` `\text(undefined)` `x→∞``lim_(x→∞)1+1/x` `=` `\text(undefined)` Finally, draw a curve along the stationary points and between but not meeting the asymptotes 
Question 5 of 5
5. Question
Graph the equation`y=(x1)e^x`Hint
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Product Rule
`y’=vu’+uv’`First, solve for the first derivative using the product rule`u` `=` `x1` `v` `=` `e^x` `u’` `=` `1` `v’` `=` `e^x` `y’` `=` `vu’+uv’` `=` `e^x(1)+(x1)e^x` Substitute known values `=` `e^x+(x1)e^x` Evaluate `=` `e^x+xe^xe^x` Distribute `=` `xe^x` `e^xe^x` cancels out Next, equate the first derivative to `0` to find the stationary point`y’=xe^x` `=` `0` `xe^x``divide e^x` `=` `0``divide e^x` Divide both sides by `e^x` `x` `=` `0` To get the value of `y`, substitute the value of `x` to the equation`y` `=` `(x1)e^x` `=` $$(\color{#CC0000}{0}1)e^{\color{#CC0000}{0}}$$ `x=0` `=` `(1)1` Anything raised to `0` is `1` `=` `1` Therefore, the stationary point will be `(0,1)`Next, determine if the stationary point would be a concave up or concave downTo do this, use the product rule on the first derivative and get the second derivative`u` `=` `x` `v` `=` `e^x` `u’` `=` `1` `v’` `=` `e^x` `y”` `=` `vu’+uv’` `=` `e^x(1)+(x)e^x` Substitute known values `=` `e^x+xe^x` Evaluate `=` `e^x(1+x)` Factorize Substitute the value of `x` to the second derivative to determine the concavity`y”` `=` `e^x(1+x)` `=` $$e^{\color{#CC0000}{0}}(1+\color{#CC0000}{0})$$ `x=0` `=` `1times1` Anything raised to `0` is `1` `=` `1` A positive concavity means that the stationary point will be a minimum, or the concave will be upwardNext, equate the second derivative to `0` to find the inflexion points`y’=e^x(1+x)` `=` `0` `e^x(1+x)``divide e^x` `=` `0``divide e^x` Divide both sides by `e^x` `1+x``1` `=` `0``1` Subtract `1` from both sides `x` `=` `1` Substitute this value of `x` to the equation to find the value of `y``y` `=` `(x1)e^x` `=` $$(\color{#CC0000}{1}1)e^{\color{#CC0000}{1}}$$ `x=0` `=` `2/e` Reciprocate `e^(1)` Therefore, the inflexion point will be `(0,2/e)``x=1` is the only xintercept, hence the curve will not be touching the xaxis on the left side