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Curve Sketching 2Curve Sketching 2
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Question 1 of 5
1. Question
Sketch the curve`f(x)=3x^4-16x^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^4-16x^3+24x^2+3` `f'(x)` `=` `12x^3-48x^2+48x=0` Equate `f'(x)` to `0` `=` `12x(x^2-4x+4)=0` Factor out `12x` `=` `4x(x-2)(x-2)=0` Factor out further `x=0,x=2` Find the corresponding `y` values by substituting each `x` value to the function`f(x)` `=` `3x^4-16x^3+24x^2+3` $$f(\color{#007DDC}{0})$$ `=` $$3(\color{#007DDC}{0}^4)-16(\color{#007DDC}{0}^3)+24(\color{#007DDC}{0}^2)+3$$ `=` `0-0+0+3` `=` `3` Stationary Point:
$$(\color{#007DDC}{0},\color{#e65021}{3})$$`f(x)` `=` `3x^4-16x^3+24x^2+3` $$f(\color{#007DDC}{2})$$ `=` $$3(\color{#007DDC}{2}^4)-16(\color{#007DDC}{2}^3)+24(\color{#007DDC}{2}^2)+3$$ `=` `48-128+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^3-48x^2+48x` `f”(x)` `=` `36x^2-96x+48` $$f”(\color{#007DDC}{0})$$ `=` $$36(\color{#007DDC}{0}^2)-96(\color{#007DDC}{0})+48$$ `=` `0-0+48` `=` `48``>``0` This means $$(\color{#007DDC}{0},\color{#e65021}{3})$$ is a minimum pointChecking $$(\color{#007DDC}{2},\color{#e65021}{19})$$:`f'(x)` `=` `12x^3-48x^2+48x` `f”(x)` `=` `36x^2-96x+48` $$f”(\color{#007DDC}{2})$$ `=` $$36(\color{#007DDC}{2}^2)-96(\color{#007DDC}{2})+48$$ `=` `144-192+48` `=` `0` This means $$(\color{#007DDC}{2},\color{#e65021}{19})$$ is an inflection point
Identify 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^3-48x^2+48x=0` $$f'(\color{#00880A}{1})$$ `=` $$12(\color{#00880A}{1})^3-48(\color{#00880A}{1}^2)+48(\color{#00880A}{1})$$ `=` `12(1)-48(1)+48` `=` `12-48+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^3-48x^2+48x=0` $$f'(\color{#00880A}{3})$$ `=` $$12(\color{#00880A}{3})^3-48(\color{#00880A}{3}^2)+48(\color{#00880A}{3})$$ `=` `12(27)-48(9)+144` `=` `324-432+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
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Question 2 of 5
2. Question
Sketch the curve`f(x)=(x^2-1)/(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^2-1)/(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^2-1)/(x^2+1)` $$f(\color{#007DDC}{0})$$ `=` $$\frac{\color{#007DDC}{0}^2-1}{\color{#007DDC}{0}^2+1}$$ `=` `-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)^2-16x(x^2+1)^2)/((x^2+1)^4)` $$f”(\color{#007DDC}{0})$$ `=` $$\frac{4(\color{#007DDC}{0}^2+1)^2-16(\color{#007DDC}{0})(\color{#007DDC}{0}^2+1)^2}{(\color{#007DDC}{0}^2+1)^4}$$ `=` `4``>``0` This means $$(\color{#007DDC}{0},\color{#e65021}{-1})$$ is a minimum point
Next, 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^2-1)/(x^2+1)` `y` `=` `(0^2-1)/(0^2+1)` Substitute `x=0` `y` `=` `-1/1` `y` `=` `-1` `x-`intercept`y` `=` `(x^2-1)/(x^2+1)` `0` `=` `(x^2-1)/(x^2+1)` Substitute `y=0` `0` `=` `x^2-1` 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^2-1)/(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
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Question 3 of 5
3. Question
Sketch the curve`f(x)=(x-4)(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)` `=` `(x-4)(x+2)^2` `f'(x)` `=` `3x^3-12=0` Equate `f'(x)` to `0` `=` `3(x^2-4)=0` Factor out `3` `=` `3(x+2)(x-2)=0` Factor out further `x=-2,x=2` Find the corresponding `y` values by substituting each `x` value to the function`f(x)` `=` `(x-4)(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)` `=` `(x-4)(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^3-12` `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^3-48x^2+48x` `f”(x)` `=` `3x^3-12` $$f”(\color{#007DDC}{2})$$ `=` $$6(\color{#007DDC}{2})$$ `=` `12``>``0` This means $$(\color{#007DDC}{2},\color{#e65021}{-32})$$ is a minimum point
Next, substitute `x=0` to the original function to find the inflection point`f(x)` `=` `(x-4)(x+2)^2` `f(0)` `=` `(0-4)(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
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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)` `=` `1-1/(x^2)=0` Equate `f'(x)` to `0` `=` `x^2-1=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` $$f(\color{#007DDC}{-1})$$ `=` $$\color{#007DDC}{-1}+\frac{1}{\color{#007DDC}{-1}}$$ `=` `-1-1` `=` `-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)` `=` `1-1/(x^2)` `=` `1-x^(-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)` `=` `1-1/(x^2)` `=` `1-x^(-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
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Question 5 of 5
5. Question
Graph the equation`y=(x-1)e^x`Hint
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Product Rule
`y’=vu’+uv’`First, solve for the first derivative using the product rule`u` `=` `x-1` `v` `=` `e^x` `u’` `=` `1` `v’` `=` `e^x` `y’` `=` `vu’+uv’` `=` `e^x(1)+(x-1)e^x` Substitute known values `=` `e^x+(x-1)e^x` Evaluate `=` `e^x+xe^x-e^x` Distribute `=` `xe^x` `e^x-e^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` `=` `(x-1)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` `=` `(x-1)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 x-intercept, hence the curve will not be touching the x-axis on the left side
