a) Function differentiates to \(f'(x)\quad =\quad 2x-4\) (careful, since it is -4, but the equation says +b, then b must be negative.

b) \(when\quad gradient\quad =\quad 0,\\ f'(x)\quad =\quad 2x-4\quad =\quad 0\\ so,\quad 2x\quad =\quad 4\\ and\quad x\quad =\quad 2\)

c) \(At\quad vertex,\quad f'(x)\quad =\quad 0\quad so\quad x\quad =\quad 2\\ Substitute\quad x\quad =\quad 2\quad into\quad f(x)={ x }^{ 2 }-4x+7\\ So,\quad y\quad =\quad 3\) (You might use your table function here)

a) Function differentiates to \(f'(x)\quad =\quad 2x+6\)

b) \(when\quad gradient\quad =\quad 0,\\ f'(x)\quad =\quad 2x+6\quad =\quad 0\\ so,\quad 2x\quad =\quad -6\\ and\quad x\quad =\quad -3\)

c) \(At\quad vertex,\quad f'(x)\quad =\quad 0\quad so\quad x\quad =\quad -3\\ Substitute\quad x\quad =\quad -3\quad into\quad f(x)={ x }^{ 2 }+6x+3\\ So,\quad y\quad =\quad -6\) (You might use your table function here)

a) \(f'(x)={ x }^{ 2 }-{ x }-12\)(Pay attention to the negative signs)

b) \(When\quad the\quad gradient\quad is\quad zero\\ f'(x)={ x }^{ 2 }-{ x }-12\quad =\quad 0\\ f'(x)=(x+3)(x-4)\quad =\quad 0\\ x\quad =\quad -3\quad or\quad 4\) You can use the polynomial solver on your GDC to solve this equation

c) Read the gradient function from your GDC for x = -3 and x = 4

a) BEWARE you are NOT differentiating here, you are just expressing it as a negative index. Remember...

\(\frac { 1 }{ { x }^{ n } } ={ x }^{ -n }\quad and\quad \frac { 3 }{ { x }^{ n } } ={ 3x }^{ -n }\quad and\quad \frac { 2 }{ { 5x }^{ n } } =\frac { 2 }{ 5 } { x }^{ -n }\) so \(f(x)={ 2x }^{ 2 }+\frac { 32 }{ x } \quad =\quad f(x)={ 2x }^{ 2 }+32{ x }^{ -1 }\)

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b) \(if,\quad f(x)={ 2x }^{ 2 }+32{ x }^{ -1 },\quad then,\quad f'(x)\quad =\quad 4x-32{ x }^{ -2 }\)

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c) \(when\quad gradient\quad =\quad 0,\\ f'(x)\quad =\quad 4x-32{ x }^{ -2 }\quad =\quad 0\\ f'(x)\quad =\quad 4x-\frac { 32 }{ { x }^{ 2 } } =0\\ so\quad 4x\quad =\quad \frac { 32 }{ { x }^{ 2 } } \\ and,\quad 4{ x }^{ 3 }\quad =\quad 32\\ and\quad { x }^{ 3 }\quad =\quad 8,\quad \\ so\quad x\quad =\quad 2\) Again, your equation solver on the GDC can do this for you.

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d) \(At\quad vertex,\quad f'(x)\quad =\quad 0\quad so\quad x\quad =\quad 2\\ Substitute\quad x\quad =\quad 2\quad into\quad f(x)={ 2x }^{ 2 }+\frac { 32 }{ x } \\ So,\quad y\quad =\quad 24\) Again, the table function will help you to avoid mistakes

a) \(if\quad V=400+3t-{ t }^{ 2 }\quad then,\frac { dV }{ dt } =3-2t\)

b) \(at\quad maximum,\quad \frac { dV }{ dt } =3-2t=0\\ so,\quad 3=2t\\ and\quad t=\frac { 3 }{ 2 } =1.5\)

c) \(Maximum\quad occurs\quad at\quad t\quad =\quad 1.5\\ when\quad t=1.5\\ V=400+3(1.5)-{ (1.5) }^{ 2 }\\ V=400+4.5-2.25\\ V=402.25,\quad V=402{ cm }^{ 3 }\quad (3sf)\)

Using the GDC, if you enter the Volume function and have the derivative switched on then you can see that at t = 1.5, Volume is 402.25 and the derivative is zero.

a) \(if\quad P(x)=19x-0.07{ x }^{ 2 }-10,\quad then,\frac { dP }{ dx } =19-0.14x\)

b) \(At\quad maximum,\quad \frac { dP }{ dx } =19-0.14x=0\\ 19=0.14x\\ x=\frac { 19 }{ 0.14 } =135.71428.....\\ x=\quad 136\quad (3sf)\)

c) \(maximum\quad profit\quad occurs\quad at\quad x\quad =\quad 136\\ Substituting\quad in\quad to\quad P(x)=19x-0.07{ x }^{ 2 }-10\\ gives\quad V=1279.2\\ V=1280\quad (3sf)\)

a) \(A=bh\\ b-h=10\quad so\\ b=10+h\quad and\\ b-10=h\\ Substitute\quad h\quad =\quad b-10\\ A=b(b-10)\\ A={ b }^{ 2 }-10b\)

b) \(if\quad A={ b }^{ 2 }-10b\\ then\quad \frac { dA }{ db } =2b-10\)

c) \(at\quad minimum\quad value,\quad \frac { dA }{ db } =2b-10=0\\ so\quad 2b=10\\ and\quad b=5\)

d) \(Substitute\quad b=5\quad into\quad A={ b }^{ 2 }-10b\\ so\quad A=-25\)

a) Substitute the value for h in to the expression for A,

\(if\quad A={ x }^{ 3 }+2xh\quad and\quad h=\frac { 3 }{ { x }^{ 3 } } \\ Substituting\quad gives\\ A={ x }^{ 3 }+\frac { 2x\times 3 }{ { x }^{ 3 } } \\ A={ x }^{ 3 }+\frac { 6x }{ { x }^{ 3 } } \\ A={ x }^{ 3 }+\frac { 6 }{ { x }^{ 2 } } \\ A={ x }^{ 3 }+6{ x }^{ -2 }\)

b) Differentiate A with respect to x

\(if\quad A={ x }^{ 3 }+6{ x }^{ -2 }\\ \\ then,\frac { dA }{ dx } =3{ x }^{ 2 }-12{ x }^{ -3 }\)

c) Solve for gradient = 0

\(At\quad minimum,\quad \frac { dA }{ dx } =3{ x }^{ 2 }-12{ x }^{ -3 }=0\\ 3{ x }^{ 2 }-12{ x }^{ -3 }=0\quad (use\quad GDC\quad solver)\\ x=1.3195....\\ x=1.32\quad (3sf)\)

a) \(Area\quad is\quad the\quad 6\quad faces\quad added\quad together\\ A\quad =\quad 2{ x }^{ 2 }+2xh+2xh\\ A=2{ x }^{ 2 }+4xh\)

b) This part is tricky but is all about making h the subject of the formula

\(A=2{ x }^{ 2 }+4xh\\ 600=2{ x }^{ 2 }+4xh\\ 600-2{ x }^{ 2 }=4xh\\ \frac { 600-2{ x }^{ 2 } }{ 4x } =h\\ \frac { 300-{ x }^{ 2 } }{ 2x } =h\)

c) This stage involves substitutuing the expression for h in to the expression for volume.

\(Since\quad V{ =x }^{ 2 }h\quad and\quad h=\frac { 300-{ x }^{ 2 } }{ 2x } \\ V=\frac { { x }^{ 2 }(300-{ x }^{ 2 }) }{ 2x } \\ V=\frac { 300{ x }^{ 2 }-{ x }^{ 4 } }{ 2x } \\ (Divide\quad through\quad by\quad 2x)\\ V=150x-\frac { 1 }{ 2 } { x }^{ 3 }\)

d) Differentiate the function and solve for gradient = 0

\(if\quad V=150x-\frac { 1 }{ 2 } { x }^{ 3 }\quad then,\frac { dV }{ dx } =150-\frac { 3 }{ 2 } { x }^{ 2 }\\ Volume\quad is\quad a\quad maximum\quad when\quad \frac { dV }{ dx } =150-\frac { 3 }{ 2 } { x }^{ 2 }=0\\ 150=\frac { 3 }{ 2 } { x }^{ 2 }\\ 300=3{ x }^{ 2 }\\ 100={ x }^{ 2 }\\ x=\quad 10\quad (-10\quad is\quad not\quad in\quad the\quad valid\quad range)\)

e) This involves substituting the value of x that gives a minimum into the expression for volume.

\(Maximum\quad occurs\quad at\quad x=\quad 10\\ substitute\quad x=10\quad into\quad V=150x-\frac { 1 }{ 2 } { x }^{ 3 }\\ V=1000{ cm }^{ 3 }\)

f) Now substitute the value of x in to the expression for h.

\(Since\quad \frac { 300-{ x }^{ 2 } }{ 2x } =h\quad and\quad x=10\\ then,\quad \frac { 300-100 }{ 20 } =h\\ 10=h\)

a) Use the formula for volume of a cylinder and rearrange.

\(V=\pi { r }^{ 2 }h\quad and\quad is\quad given\quad as\quad 400{ cm }^{ 3 }\\ then\\ 400=\pi { r }^{ 2 }h\\ and\quad so,\quad h=\frac { 400 }{ \pi { r }^{ 2 } } \)

b) Use the formula for surface area of a cylinder and substitute the expression for h, then simplify.

\(A=2\pi { r }^{ 2 }+2\pi { r }h\quad and\quad h=\frac { 400 }{ \pi { r }^{ 2 } } \\ Substituting\quad for\quad h,\\ A=2\pi { r }^{ 2 }+\frac { 2\pi { r }\times 400 }{ \pi { r }^{ 2 } } \\ Cancelling\quad common\quad factors\\ A=2\pi { r }^{ 2 }+\frac { 800 }{ { r } } \\ A=2\pi { r }^{ 2 }+800{ r }^{ -1 }\)

c) Solve for gradient = 0. Perfectly acceptable to use the solve function to do this.

\(If\quad A=2\pi { r }^{ 2 }+800{ r }^{ -1 },\quad then\quad \frac { dA }{ dr } =4\pi { r }-800{ r }^{ -2 }\\ At\quad minimum,\quad \frac { dA }{ dr } =4\pi { r }-800{ r }^{ -2 }=0\\ 4\pi { r }=800{ r }^{ -2 }\\ 4\pi { r }=\frac { 800 }{ { r }^{ 2 } } \\ 4\pi { r }^{ 3 }=800\\ { r }^{ 3 }=\frac { 800 }{ 4\pi } \\ r=\sqrt [ 3 ]{ \frac { 800 }{ 4\pi } } \\ r=3.9929454...\\ r=3.99\quad cm\quad (3sf)\)

d) Substitutions can be done using the table function on your GDC or otherwise.

i) \(Substitute\quad r=3.99\quad in\quad to\quad h=\frac { 400 }{ \pi { r }^{ 2 } } \\ h=7.997685...\\ h=8.00cm\quad (3sf)\)

ii) \(Substitute\quad r=3.99\quad in\quad to\quad A=2\pi { r }^{ 2 }+800{ r }^{ -1 }\\ A=300.5301915...\\ A=\quad 301{ cm }^{ 2 }\quad (3sf)\)