Bepaal $f'(x)$ vanuit eerste beginels indien $f(x) = -4x^2$
Bepaal:
7.2.1 $f'(x)$ indien $f(x) = 2x^3 - 3x$
7.2.2 $D_rig( rac{1}{
oot{3}{x^2} + 2x^5}ig)$
Vir watter waardes van $x$ sal die raaklyn aan $f(x) = -2x^2 + 8x$ 'n positieve helling hê? - NSC Mathematics - Question 7 - 2023 - Paper 1
Question 7
Bepaal $f'(x)$ vanuit eerste beginels indien $f(x) = -4x^2$
Bepaal:
7.2.1 $f'(x)$ indien $f(x) = 2x^3 - 3x$
7.2.2 $D_rig( rac{1}{
oot{3}{x^2} + 2x^5}ig)$
Vir w... show full transcript
Worked Solution & Example Answer:Bepaal $f'(x)$ vanuit eerste beginels indien $f(x) = -4x^2$
Bepaal:
7.2.1 $f'(x)$ indien $f(x) = 2x^3 - 3x$
7.2.2 $D_rig( rac{1}{
oot{3}{x^2} + 2x^5}ig)$
Vir watter waardes van $x$ sal die raaklyn aan $f(x) = -2x^2 + 8x$ 'n positieve helling hê? - NSC Mathematics - Question 7 - 2023 - Paper 1
Step 1
Bepaal $f'(x)$ vanuit eerste beginels indien $f(x) = -4x^2$
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the derivative f′(x), we apply the definition of the derivative:
f'(x) = rac{f(x+h) - f(x)}{h}
Substituting f(x)=−4x2:
f'(x) = rac{-4(x+h)^2 - (-4x^2)}{h}
This simplifies to:
f'(x) = rac{-4(x^2 + 2xh + h^2) + 4x^2}{h} = rac{-8xh - 4h^2}{h}
Further simplification gives:
f′(x)=−8x
Thus, the derivative is:
f′(x)=−8x
Step 2
Bepaal: 7.2.1 $f'(x)$ indien $f(x) = 2x^3 - 3x$
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the domain, we need to ensure the denominator is not zero:
oot{3}{x^2} + 2x^5
eq 0$$
Setting the equation to zero, we find the critical points and test intervals for valid solutions.
This leads to critical values where the expression is defined excluding $x = 0$ and other potential roots.
Step 4
Vir watter waardes van $x$ sal die raaklyn aan $f(x) = -2x^2 + 8x$ 'n positieve helling hê?
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
First, we determine the derivative:
f′(x)=−4x+8
To find where the slope is positive, we set:
−4x+8>0
Solving this inequality gives:
x<2
Therefore, the values for x where the tangent line has a positive slope are:
x<2
