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Parents Pricing Home NSC Mathematics Finance growth and decay 7.1 Determine $f’(x)$ from first principles if $f(x)=x^{2}+x.$
7.2 Determine $f’(x)$ if $f(v)=2v^{2}-3v^{3}+8v$
7.1 Determine $f’(x)$ from first principles if $f(x)=x^{2}+x.$
7.2 Determine $f’(x)$ if $f(v)=2v^{2}-3v^{3}+8v$ - NSC Mathematics - Question 7 - 2022 - Paper 1 Question 7
View full question 7.1 Determine $f’(x)$ from first principles if $f(x)=x^{2}+x.$
7.2 Determine $f’(x)$ if $f(v)=2v^{2}-3v^{3}+8v$.
7.3 The tangent to $g(x)=ax^{4}+3x^{3}+bx+c$ has a... show full transcript
View marking scheme Worked Solution & Example Answer:7.1 Determine $f’(x)$ from first principles if $f(x)=x^{2}+x.$
7.2 Determine $f’(x)$ if $f(v)=2v^{2}-3v^{3}+8v$ - NSC Mathematics - Question 7 - 2022 - Paper 1
Determine $f’(x)$ from first principles if $f(x)=x^{2}+x.$ Only available for registered users.
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To find the derivative using first principles, we apply the limit definition:
f ’ ( x ) = e x t l i m h o 0 f ( x + h ) − f ( x ) h f’(x) = ext{lim}_{h o 0} \frac{f(x+h) - f(x)}{h} f ’ ( x ) = e x t l im h o 0 h f ( x + h ) − f ( x )
Given that f ( x ) = x 2 + x f(x) = x^{2} + x f ( x ) = x 2 + x
f ( x + h ) = ( x + h ) 2 + ( x + h ) = x 2 + 2 x h + h 2 + x + h f(x+h) = (x+h)^{2} + (x+h) = x^{2} + 2xh + h^{2} + x + h f ( x + h ) = ( x + h ) 2 + ( x + h ) = x 2 + 2 x h + h 2 + x + h
Therefore, f ( x + h ) − f ( x ) = ( x 2 + 2 x h + h 2 + x + h ) − ( x 2 + x ) = 2 x h + h 2 + h f(x+h) - f(x) = (x^{2} + 2xh + h^{2} + x + h) - (x^{2} + x) = 2xh + h^{2} + h f ( x + h ) − f ( x ) = ( x 2 + 2 x h + h 2 + x + h ) − ( x 2 + x ) = 2 x h + h 2 + h
Now substituting into the limit:
f ’ ( x ) = e x t l i m h o 0 2 x h + h 2 + h h f’(x) = ext{lim}_{h o 0} \frac{2xh + h^{2} + h}{h} f ’ ( x ) = e x t l im h o 0 h 2 x h + h 2 + h
Canceling h h h
f ’ ( x ) = e x t l i m h o 0 ( 2 x + h + 1 ) f’(x) = ext{lim}_{h o 0} (2x + h + 1) f ’ ( x ) = e x t l im h o 0 ( 2 x + h + 1 )
Taking the limit as h h h 0 0 0
f ’ ( x ) = 2 x + 1. f’(x) = 2x + 1. f ’ ( x ) = 2 x + 1.
This means that f ’ ( x ) = 2 x + 1 f’(x) = 2x + 1 f ’ ( x ) = 2 x + 1
Determine $f’(x)$ if $f(v)=2v^{2}-3v^{3}+8v$. Only available for registered users.
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To find f ’ ( v ) f’(v) f ’ ( v )
Differentiate the function term by term:
The derivative of 2 v 2 2v^{2} 2 v 2 4 v 4v 4 v
The derivative of − 3 v 3 -3v^{3} − 3 v 3 − 9 v 2 -9v^{2} − 9 v 2
The derivative of 8 v 8v 8 v 8 8 8
Therefore,
f ’ ( v ) = 4 v − 9 v 2 + 8. f’(v) = 4v - 9v^{2} + 8. f ’ ( v ) = 4 v − 9 v 2 + 8.
Arrange it neatly:
f ’ ( v ) = − 9 v 2 + 4 v + 8. f’(v) = -9v^{2} + 4v + 8. f ’ ( v ) = − 9 v 2 + 4 v + 8.
The tangent to $g(x)=ax^{4}+3x^{3}+bx+c$ has a minimum gradient at the point $(-1;-7)$. For which values of $x$ will $g$ be concave up? Only available for registered users.
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To determine where the function g ( x ) g(x) g ( x )
First, compute the first derivative:
g ’ ( x ) = 4 a x 3 + 9 x 2 + b . g’(x) = 4ax^{3} + 9x^{2} + b. g ’ ( x ) = 4 a x 3 + 9 x 2 + b .
Next, compute the second derivative:
g ’’ ( x ) = 12 a x 2 + 18 x . g’’(x) = 12ax^{2} + 18x. g ’’ ( x ) = 12 a x 2 + 18 x .
For the function to be concave up, we require:
g ’’ ( x ) > 0. g’’(x) > 0. g ’’ ( x ) > 0.
Analyze at the point x = − 1 x = -1 x = − 1
Substituting x = − 1 x = -1 x = − 1 g ’ ( − 1 ) = 6 a ( − 1 ) + 6 = 0. g’(-1) = 6a(-1) + 6 = 0. g ’ ( − 1 ) = 6 a ( − 1 ) + 6 = 0.
For concavity, if we simplify:
ightarrow 12a - 18 > 0.$$
This leads to:
a > 1. a > 1. a > 1. g ( x ) g(x) g ( x ) x x x a > 1 a > 1 a > 1
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